Q: If the latus-rectum of an ellipse is one half of its minor axis, then its eccentricity is

Consider the equation of the ellipse is \$\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}=1\$It is given that; length of latus rectum = half of minor axis\$\dfrac {2b^2}{a}=b \Rightarrow a=2b\$now, \$b^2=a^2 (1-e^2)\$\$\Rightarrow \ b^2=4b^2 (1-e^2)\Rightarrow 1-e^2 =1/4 \Rightarrow e^2 =3/4\$\$\therefore \ \$ eccentricity \$(e)=\sqrt 3 /2\$

Q: If major axis is the x-axis and passes through the points \$(4, 3)\$ and \$(6, 2)\$, then the equation for the ellipse whose centre is the origin is satisfies the given condition.

Major axis is x - axisLet the equation of the ellipse be \$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} =1\$\$(4,3)\$ and \$(6, 2)\$ lies on it \$\displaystyle \frac{16}{a^2} + \frac{9}{b^2} =1\$ ........ \$(i)\$\$\displaystyle \frac{36}{a^2} + \frac{4}{b^2} =1\$ ........ \$(ii)\$Subtracting \$(ii)\$ from \$(i)\$, we get \$\displaystyle \frac{-20}{a^2} + \frac{5}{b^2} =0\$ \$\Rightarrow 5a^2 = 20 b^2 \Rightarrow a^2 = 4b^2\$Putting the value of \$a^2\$ in eqn. \$(i)\$ \$\displaystyle \frac{16}{4b^2} + \frac{9}{b^2} = 1\$\$\therefore b^2 =13\$ and \$a^2 = 4b^2 = 4 \times 13 =52\$\$\therefore \$ Equation of ellipse is \$\displaystyle \frac{x^2 }{52}+\frac{y^2}{13}=1\$

Q: The equation of the ellipse whose foci are \$(\pm5,0)\$ and of the directrix is \$5x=36\$, is

Given foci \$(\pm 5,0)\$ and directrix \$x=\cfrac{36}{5}\$Then \$ae=5\$ (focus coordinates (\$\pm ae,0)]\$....(1)\$\cfrac{a}{e}=\cfrac{36}{5}\$ (directrix equation \$x=\cfrac{a}{e}\$]....(2)From (1) and (2) \${a}^{2}=36\Rightarrow\$ \$a=6\$\$e=\cfrac{5}{6}\Rightarrow \$ \$\sqrt { 1-\cfrac { { b }^{ 2 } }{ { a }^{ 2 } } } =\cfrac { 5 }{ 6 } \$\$1-\cfrac { { b }^{ 2 } }{ 36 } =\cfrac{25}{36}\$\$b=\sqrt 11\$required equation \$\cfrac{{x}^{2}}{36}+\cfrac{{y}^{2}}{11}=1\$

Q: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \$\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{100}= 1\$

The given equation is \$\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{100}= 1\$ or \$\displaystyle \frac{x^{2}}{5^{2}}+\frac{y^{2}}{10^{2}}=1\$Here, the denominator of \$\displaystyle \frac{y^{2}}{100}\$ is greater than the denominator of \$\displaystyle \frac{x^{2}}{25}.\$Therefore the major axis is along the \$y\$-axis while the minor axis is along the \$x\$-axis. On comparing the given equation with \$\displaystyle \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}= 1\$ we obtain \$b = 5\$ and \$a = 10\$\$\displaystyle \therefore ae=c =\sqrt{a^{2}-b^{2}}=\sqrt{100-25}=\sqrt{75}=5\sqrt{3}\$The coordinates of the foci are \$\displaystyle \left ( 0, \pm 5\sqrt{3} \right )\$The coordinates of the vertices are \$\displaystyle \left ( 0, \pm 10 \right )\$Length of major axis \$= 2a = 20\$Length of minor axis \$= 2b = 10\$Eccentricity \$e =\$ \$\displaystyle \frac{c}{a}= \frac{5\sqrt{3}}{10}= \frac{\sqrt{3}}{2}\$Length of latus rectum \$=\$ \$\displaystyle \frac{2b^{2}}{a}= \frac{2\times 25}{10}= 5\$

Q: If the eccentricity of the hyperbola \${ x }^{ 2 }-{ y }^{ 2 }\sec ^{ 2 }{ \alpha } =5\$ is \$\sqrt { 3 } \$ times the eccentricity of the ellipse \${ x }^{ 2 }\sec ^{ 2 }{ \alpha } +{ y }^{ 2 }=25\$, then the value of \$\alpha \$ is

Equation for ellipse can be written as\$\dfrac{x^2}{25\cos ^2\alpha}+\dfrac{y^2}{25}=1\$the eccentricity of the given ellipse \$e_e^2=1-\dfrac{25\cos^2\alpha}{25}=\sin^2\alpha\$And the eccentricity of the Hyperbola is given as \$e_e^2=1+\dfrac{5\cos^2\alpha}{5}=1+\cos^2\alpha\$\$e_h^2=3e_e^2\$ (given in the question)\$1+\cos^2\alpha=3\sin^2\alpha\\ \sin^2\alpha=1/2\\ \alpha=\dfrac{\pi}{4}\$

Q: The focus of an ellipse is at the origin. The directrix is the line \$\displaystyle x= 4\$ and its eccentricity is \$\displaystyle \dfrac12\$ then length of its semi major axis is

Given directrix is \$x=4\$ Eccentricity\$=\dfrac{1}2\$ Focus is at \$(0,0)\$Shifting focus to \$(h,0)\$ such that centre becomes \$(0,0).\$Equation of directrix becomes \$x=4+h\$ which gives Distance between center and focus i.e, \$ ae=h \$ and Distance between center and directrix i.e, \$\dfrac{a}e=4+h\$Solving for \$a\$\$e=\dfrac{1}2 \Rightarrow a=2h \$ and \$ 2a=4+h\$\$2a=4+\dfrac{a}2 \Rightarrow a=\dfrac{8}3.\$

Q: Find the eccentricity, foci, length of the Latus rectum and the equations of directrices of the ellipse \$9x^2 + 16y^2 = 144\$.

Given equation can be written in standard form as \$\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\$ \$\Rightarrow a^2=16, b^2=9\Rightarrow a=4, b=3\$So eccentricity \$,e=\sqrt{1-\dfrac{b^2}{a^2}}=\sqrt{1-\dfrac{9}{16}}=\dfrac{\sqrt 7}{4}\$Thus foci \$=(\pm ae,0)=(\pm \sqrt 7,0)\$Length of latus rectum \$=\dfrac{2b^2}{a}=\dfrac{2\cdot 9}{4}=\dfrac{9}{2}\$And equation of directrices are \$x=\pm \dfrac{a}{e}=\pm \dfrac{16}{\sqrt 7}\$

Q: A rod of length \$12\$ cm moves with it ends always touching the coordinate axes. Dertermine the equation of the locus of a point \$P\$ on the rod which is \$3\$ cm from the end in contact with the \$x\$-axis

Let \$AB\$ be the rod making an angle \$\displaystyle \theta \$ with positive direction of \$x\$-axis and \$P (x, y)\$ be the point on it such that \$AP= 3cm\$Now, \$ PB = AB - AP = (12 - 3) cm = 9 cm \$ \$\displaystyle \left (AB = 12 cm \right)\$ Draw \$\displaystyle PQ \perp OY\$ and \$\displaystyle PR\perp OX\$In \$\displaystyle \bigtriangleup PBQ \$, \$\displaystyle \cos \theta =\dfrac{PQ}{PB}= \dfrac{x}{9}\$In \$\displaystyle \bigtriangleup PRA\$, \$\displaystyle \sin \theta =\frac{PR}{PA}= \frac{y}{3}\$Since \$\displaystyle \sin ^{2}\theta +\cos ^{2}\theta = 1\$\$\Rightarrow \displaystyle \left ( \dfrac{y}{3} \right )^{2}+\left ( \dfrac{x}{9} \right )^{2}= 1\$\$\Rightarrow \displaystyle \dfrac{x^{2}}{81}+\dfrac{y^{2}}{9}= 1\$Thus the equation of the locus of point \$P\$ on the rod is \$\displaystyle \dfrac{x^{2}}{81}+\dfrac{y^{2}}{9}= 1\$

Q: If \$\left( 5,12 \right) \$ and \$\left( 24,7 \right) \$ are the foci of an ellipse passing through the origin, then the eccentricity of the conic is

Let \$S\left( 5,12 \right),S'\left( 24,7 \right) \$ be the two foci.\$P(0,0)\$ is a point on the conic. \$SP=\sqrt { 25+144 } =13\$\$S'P=\sqrt { 576+49 } =\sqrt { 625 } =25\$\$SS'=\sqrt { { \left( 24-5 \right) }^{ 2 }+{ \left( 7-12 \right) }^{ 2 } } =\sqrt { { 19 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 386 } \$If the conic is an ellipse, then\$SP+S'P=2a\$ and \$SS'=2ae\$\$\displaystyle \therefore e=\frac { SS' }{ SP+S'P } =\frac { \sqrt { 386 } }{ 13+25 } =\frac { \sqrt { 386 } }{ 38 } \$

Question 1

If the latus-rectum of an ellipse is one half of its minor axis, then its eccentricity is

Accepted Answer

Consider the equation of the ellipse is $\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}=1$It is given that; length of latus rectum = half of minor axis$\dfrac {2b^2}{a}=b \Rightarrow a=2b$now, $b^2=a^2 (1-e^2)$$\Rightarrow \ b^2=4b^2 (1-e^2)\Rightarrow 1-e^2 =1/4 \Rightarrow e^2 =3/4$$\therefore \ $ eccentricity $(e)=\sqrt 3 /2$

Question 2

If major axis is the x-axis and passes through the points $(4, 3)$ and $(6, 2)$, then the equation for the ellipse whose centre is the origin is satisfies the given condition.

Accepted Answer

Major axis is x - axisLet the equation of the ellipse be $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} =1$$(4,3)$ and $(6, 2)$ lies on it $\displaystyle \frac{16}{a^2} + \frac{9}{b^2} =1$ ........ $(i)$$\displaystyle \frac{36}{a^2} + \frac{4}{b^2} =1$ ........ $(ii)$Subtracting $(ii)$ from $(i)$, we get $\displaystyle \frac{-20}{a^2} + \frac{5}{b^2} =0$ $\Rightarrow 5a^2 = 20 b^2  \Rightarrow a^2 = 4b^2$Putting the value of $a^2$ in eqn. $(i)$ $\displaystyle \frac{16}{4b^2} + \frac{9}{b^2} = 1$$\therefore b^2 =13$ and $a^2 = 4b^2 = 4 \times 13 =52$$\therefore $ Equation of ellipse is $\displaystyle \frac{x^2 }{52}+\frac{y^2}{13}=1$

Question 3

The equation of the ellipse whose foci are $(\pm5,0)$ and of the directrix is $5x=36$, is

Accepted Answer

Given foci $(\pm 5,0)$ and directrix $x=\cfrac{36}{5}$Then $ae=5$ (focus coordinates ($\pm ae,0)]$....(1)$\cfrac{a}{e}=\cfrac{36}{5}$ (directrix equation $x=\cfrac{a}{e}$]....(2)From (1) and (2) ${a}^{2}=36\Rightarrow$ $a=6$$e=\cfrac{5}{6}\Rightarrow $ $\sqrt { 1-\cfrac { { b }^{ 2 } }{ { a }^{ 2 } }  } =\cfrac { 5 }{ 6 } $$1-\cfrac { { b }^{ 2 } }{ 36 } =\cfrac{25}{36}$$b=\sqrt 11$required equation $\cfrac{{x}^{2}}{36}+\cfrac{{y}^{2}}{11}=1$

Question 4

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{100}= 1$

Accepted Answer

The given equation is $\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{100}= 1$ or $\displaystyle \frac{x^{2}}{5^{2}}+\frac{y^{2}}{10^{2}}=1$Here, the denominator of $\displaystyle \frac{y^{2}}{100}$ is greater than the denominator of $\displaystyle \frac{x^{2}}{25}.$Therefore the major axis is along the $y$-axis while the minor axis is along the $x$-axis. On comparing the given equation with $\displaystyle \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}= 1$ we obtain $b = 5$ and $a = 10$$\displaystyle \therefore ae=c =\sqrt{a^{2}-b^{2}}=\sqrt{100-25}=\sqrt{75}=5\sqrt{3}$The coordinates of the foci are $\displaystyle \left ( 0, \pm 5\sqrt{3} \right )$The coordinates of the vertices are $\displaystyle \left ( 0, \pm 10 \right )$Length of major axis $= 2a = 20$Length of minor axis $= 2b = 10$Eccentricity $e =$ $\displaystyle \frac{c}{a}= \frac{5\sqrt{3}}{10}= \frac{\sqrt{3}}{2}$Length of latus rectum $=$ $\displaystyle \frac{2b^{2}}{a}= \frac{2\times 25}{10}= 5$

Question 5

If the eccentricity of the hyperbola ${ x }^{ 2 }-{ y }^{ 2 }\sec ^{ 2 }{ \alpha  } =5$ is $\sqrt { 3 } $ times the eccentricity of the ellipse ${ x }^{ 2 }\sec ^{ 2 }{ \alpha  } +{ y }^{ 2 }=25$, then the value of $\alpha $ is

Accepted Answer

Equation for ellipse can be written as$\dfrac{x^2}{25\cos ^2\alpha}+\dfrac{y^2}{25}=1$the eccentricity of the given ellipse $e_e^2=1-\dfrac{25\cos^2\alpha}{25}=\sin^2\alpha$And the eccentricity  of the Hyperbola is given as $e_e^2=1+\dfrac{5\cos^2\alpha}{5}=1+\cos^2\alpha$$e_h^2=3e_e^2$ (given in the question)$1+\cos^2\alpha=3\sin^2\alpha\ \sin^2\alpha=1/2\ \alpha=\dfrac{\pi}{4}$

Question 6

The focus of an ellipse is at the origin. The directrix is the line $\displaystyle x= 4$ and its eccentricity is $\displaystyle \dfrac12$ then length of its semi major axis is

Accepted Answer

Given directrix is $x=4$ Eccentricity$=\dfrac{1}2$ Focus is at $(0,0)$Shifting focus to $(h,0)$ such that centre becomes  $(0,0).$Equation of directrix becomes $x=4+h$ which gives Distance between center and focus i.e, $ ae=h $ and Distance between center and directrix i.e, $\dfrac{a}e=4+h$Solving for $a$$e=\dfrac{1}2 \Rightarrow  a=2h $ and $  2a=4+h$$2a=4+\dfrac{a}2 \Rightarrow a=\dfrac{8}3.$

Question 7

Find the eccentricity, foci, length of the Latus rectum and the equations of directrices of the ellipse $9x^2 + 16y^2 = 144$.

Accepted Answer

Given equation can be written in standard form as $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ $\Rightarrow a^2=16, b^2=9\Rightarrow a=4, b=3$So eccentricity $,e=\sqrt{1-\dfrac{b^2}{a^2}}=\sqrt{1-\dfrac{9}{16}}=\dfrac{\sqrt 7}{4}$Thus foci $=(\pm ae,0)=(\pm \sqrt 7,0)$Length of latus rectum $=\dfrac{2b^2}{a}=\dfrac{2\cdot 9}{4}=\dfrac{9}{2}$And equation of directrices are $x=\pm \dfrac{a}{e}=\pm \dfrac{16}{\sqrt 7}$

Question 8

Find the eccentricity, foci and the length of the latus-rectum of the ellipse $x^{2}+4y^{2}+8y-2x+1=0$.

Accepted Answer

Find eccentricity, foci & length of latus - rectum of ellipse$ \therefore x^{2}+4y^{2}+8y-2x+1 = 0 $$ \therefore x^{2}-2x+1+4y^{2}+8y+2-1 = 0$$ (x-1)^{2}+(2y+2)^{2}-2 = 0$$ \therefore (x-1)^{2}+(2y+2)^{2} = 2$$ \therefore \frac{(x-1)^{2}}{2}+(2)^{2}\frac{(y+1)^{2}}{2} = 1 $$ \therefore \frac{(x-1)^{2}}{2}+\frac{(y+1)^{2}}{1/2} = 1 $$ \rightarrow x_{1} = 1 $ & $ y_{1}=-1 $Also,$ a = \sqrt{2}$ &$ b = \frac{1}{\sqrt{2}}$centre $ = (1,-3)$$ \rightarrow $ majer axis$ = 2a = 2\times \sqrt{2}=2\sqrt{2}$$ \rightarrow $ minor axis$ = 2b = 2\times \frac{1}{\sqrt{2}}=\sqrt{2}$$ \rightarrow e = \sqrt{1-\frac{b^{2}}{a^{2}}}$$ e = \sqrt{1-\frac{1/2}{2}}=\sqrt{1-\frac{1}{4}} = \sqrt[0]{\frac{3}{4}}  = \frac{\sqrt{3}}{2}$$ \rightarrow $ foci $ = (x_{1}\pm ae,y_{1}) = (1\pm \sqrt{2}.\frac{\sqrt{3}}{2},-1)$$ = (1\pm \sqrt{\frac{3}{2}},-1)$$ \rightarrow$ length of latus $ = \frac{2b^{2}}{a} = \frac{2\times 1/2}{2} = \frac{1}{2}$ Ans

Question 9

A rod of length $12$ cm moves with it ends always touching the coordinate axes. Dertermine the equation of the locus of a point $P$ on the rod which is $3$ cm from the end in contact with the $x$-axis

Accepted Answer

Let $AB$ be the rod making an angle $\displaystyle \theta $ with positive direction of $x$-axis and $P (x, y)$ be the point on it such that $AP= 3cm$Now, $ PB = AB - AP = (12 - 3) cm = 9 cm $   $\displaystyle \left (AB = 12 cm \right)$ Draw $\displaystyle PQ \perp OY$ and $\displaystyle PR\perp OX$In $\displaystyle \bigtriangleup PBQ $, $\displaystyle \cos \theta =\dfrac{PQ}{PB}= \dfrac{x}{9}$In $\displaystyle \bigtriangleup PRA$, $\displaystyle \sin  \theta =\frac{PR}{PA}= \frac{y}{3}$Since $\displaystyle \sin ^{2}\theta +\cos ^{2}\theta = 1$$\Rightarrow \displaystyle \left ( \dfrac{y}{3} \right )^{2}+\left ( \dfrac{x}{9} \right )^{2}= 1$$\Rightarrow \displaystyle \dfrac{x^{2}}{81}+\dfrac{y^{2}}{9}= 1$Thus the equation of the locus of point $P$ on the rod is $\displaystyle \dfrac{x^{2}}{81}+\dfrac{y^{2}}{9}= 1$

Question 10

If $\left( 5,12 \right) $ and $\left( 24,7 \right) $ are the foci of an ellipse passing through the origin, then the eccentricity of the conic is

Accepted Answer

Let $S\left( 5,12 \right),S'\left( 24,7 \right) $ be the two foci.$P(0,0)$ is a point on the conic. $SP=\sqrt { 25+144 } =13$$S'P=\sqrt { 576+49 } =\sqrt { 625 } =25$$SS'=\sqrt { { \left( 24-5 \right)  }^{ 2 }+{ \left( 7-12 \right)  }^{ 2 } } =\sqrt { { 19 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 386 } $If the conic is an ellipse, then$SP+S'P=2a$ and $SS'=2ae$$\displaystyle \therefore e=\frac { SS' }{ SP+S'P } =\frac { \sqrt { 386 }  }{ 13+25 } =\frac { \sqrt { 386 }  }{ 38 } $

Revise with Concepts

Introduction to Ellipse

Standard Equation of Ellipse

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Quick Summary With Stories

Equation of an ellipse

If the latus-rectum of an ellipse is one half of its minor axis, then its eccentricity is

If major axis is the x-axis and passes through the points $(4, 3)$ and $(6, 2)$ , then the equation for the ellipse whose centre is the origin is satisfies the given condition.

The equation of the ellipse whose foci are $(\pm 5, 0)$ and of the directrix is $5 x = 36$ , is

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x ^{2}}{2 5} + \frac{y ^{2}}{1 0 0} = 1$

If the eccentricity of the hyperbola $x^{2} - y^{2} sec^{2} α = 5$ is $3$ times the eccentricity of the ellipse $x^{2} sec^{2} α + y^{2} = 25$ , then the value of $α$ is

The focus of an ellipse is at the origin. The directrix is the line $x = 4$ and its eccentricity is $\frac{1}{2}$ then length of its semi major axis is

Find the eccentricity, foci, length of the Latus rectum and the equations of directrices of the ellipse $9 x^{2} + 16 y^{2} = 144$ .

Find the eccentricity, foci and the length of the latus-rectum of the ellipse $x^{2} + 4 y^{2} + 8 y - 2 x + 1 = 0$ .

A rod of length $12$ cm moves with it ends always touching the coordinate axes. Dertermine the equation of the locus of a point $P$ on the rod which is $3$ cm from the end in contact with the $x$ -axis

If $(5, 12)$ and $(24, 7)$ are the foci of an ellipse passing through the origin, then the eccentricity of the conic is

Revise with Concepts

Introduction to Ellipse

Standard Equation of Ellipse

Learn with Videos

Quick Summary With Stories

Equation of an ellipse

If the latus-rectum of an ellipse is one half of its minor axis, then its eccentricity is

If major axis is the x-axis and passes through the points (4,3) and (6,2), then the equation for the ellipse whose centre is the origin is satisfies the given condition.

The equation of the ellipse whose foci are (±5,0) and of the directrix is 5x=36, is

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 25x2​+100y2​=1

If the eccentricity of the hyperbola x2−y2sec2α=5 is 3​ times the eccentricity of the ellipse x2sec2α+y2=25, then the value of α is

The focus of an ellipse is at the origin. The directrix is the line x=4 and its eccentricity is 21​ then length of its semi major axis is

Find the eccentricity, foci, length of the Latus rectum and the equations of directrices of the ellipse 9x2+16y2=144.

Find the eccentricity, foci and the length of the latus-rectum of the ellipse x2+4y2+8y−2x+1=0.

A rod of length 12 cm moves with it ends always touching the coordinate axes. Dertermine the equation of the locus of a point P on the rod which is 3 cm from the end in contact with the x-axis

If (5,12) and (24,7) are the foci of an ellipse passing through the origin, then the eccentricity of the conic is

If major axis is the x-axis and passes through the points $(4, 3)$ and $(6, 2)$ , then the equation for the ellipse whose centre is the origin is satisfies the given condition.

The equation of the ellipse whose foci are $(\pm 5, 0)$ and of the directrix is $5 x = 36$ , is

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x ^{2}}{2 5} + \frac{y ^{2}}{1 0 0} = 1$

If the eccentricity of the hyperbola $x^{2} - y^{2} sec^{2} α = 5$ is $3$ times the eccentricity of the ellipse $x^{2} sec^{2} α + y^{2} = 25$ , then the value of $α$ is

The focus of an ellipse is at the origin. The directrix is the line $x = 4$ and its eccentricity is $\frac{1}{2}$ then length of its semi major axis is

Find the eccentricity, foci, length of the Latus rectum and the equations of directrices of the ellipse $9 x^{2} + 16 y^{2} = 144$ .

Find the eccentricity, foci and the length of the latus-rectum of the ellipse $x^{2} + 4 y^{2} + 8 y - 2 x + 1 = 0$ .

A rod of length $12$ cm moves with it ends always touching the coordinate axes. Dertermine the equation of the locus of a point $P$ on the rod which is $3$ cm from the end in contact with the $x$ -axis

If $(5, 12)$ and $(24, 7)$ are the foci of an ellipse passing through the origin, then the eccentricity of the conic is