Tag: tangent and normal to an ellipse

Questions Related to tangent and normal to an ellipse

If a normal is drawn at point $P$ of ellipse $ \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, then the maximum distance from centre of ellipse will be $a-b$ 

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. True

  2. False

Show answer
Correct Option: A

If the normal at any point $P$ of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ meets the axes in $G$ and $g$ respectively, then $|PG| : |Pg|$ is equal to 

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $a:b$

  2. $a^2:b^2$

  3. $b^2:a^2$

  4. $b:a$

Show answer
Correct Option: A

One foot of normal of the ellipse $4x^2$ $+$ 9$y^2$ $= 36 $, that is parallel to the line $2x + y = 3 $, is

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $\left ( \dfrac{9}{8}, \dfrac{5}{8} \right )$

  2. $\left ( \dfrac{9}{8}, \dfrac{8}{5} \right )$

  3. $\left ( \dfrac{8}{9}, \dfrac{8}{5} \right )$

  4. None

Show answer
Correct Option: B

If the normal at any point on the ellipse $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ meets the axes in $G$ and $g$ respectively, then $PG:Pg=$

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $a:b$

  2. ${ a }^{ 2 }:{ b }^{ 2 }$

  3. $b:a$

  4. ${ b }^{ 2 }:{ a }^{ 2 }$

Show answer
Correct Option: A

The equation of normal at the point $(0, 3)$ of the ellipse $9x^2 + 5y^2 = 45$ is

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $y-  3 = 0$

  2. $y + 3 = 0$

  3. $x$-axis

  4. $y$-axis

Show answer
Correct Option: C
Explanation:

$9x^2+5y^2=45$


Differentiating with respect to $x$, we get,

$18x+10y\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=-\dfrac{18x}{10y}$

Therefore, Slope of tangent at $(0,3)=\dfrac{-0}{30}=0$

Slope of normal = $\dfrac{1}{0}$

The equation of normal is :
$y-3=\dfrac{1}{0}(x-0)$
$x-0=0$
$x=0$

Find the equation of the normal to the ellipse $9x^2 + 16y^2 = 288$ at the point $(4, 3).$

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $4x-3y=7$

  2. $3x-4y=7$

  3. $4x+3y=7$

  4. $3x+4y=7$

Show answer
Correct Option: A
Explanation:

Given ellipse may be written as, $\displaystyle \frac{x^2}{32}+\frac{y^2}{18}=1$
$\Rightarrow a^2 = 32, b^2 = 18$
Hence required normal at $(4,3)$ is given by,
$\displaystyle y-3=\frac{3\times 32}{4\times 18}(x-4)\Rightarrow y-3=\frac{4}{3}(x-4)\Rightarrow 4x-3y=7$

The line $y=mx-\dfrac{\left(a^{2}-b^{2}\right)m}{\sqrt{a^{2}b^{2}m^{2}}}$ is normal to  the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ for all values of $m$ belongs to 

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $\left(0,1\right)$

  2. $\left(0,\infty\right)$

  3. $R$

  4. $none\ of\ these$

Show answer
Correct Option: A

The number of normals to the ellipse $\dfrac { { x }^{ 2 } }{ 25 } +\dfrac { { y }^{ 2 } }{ 16 } =1$ which are tangents to the circle ${ x }^{ 2 }+{ y }^{ 2 }=9$ is

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. 1

  2. 2

  3. 3

  4. 0

Show answer
Correct Option: A

The equation of the normal to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ at the end of latus rectum in quadrant $1^{st}$ and $4^{th}$ is

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $x-ey-ae^3=0$

  2. $x+ey-ae^3=0$

  3. $y-ex-be^3=0$

  4. $y+ex-be^3=0$

Show answer
Correct Option: A,B
Explanation:

Ellipse:$\cfrac { { x }^{ 2 } }{ a^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b^{ 2 } } } =$

ends of L.R in 1st and 4th quadrant is $L(ae,\cfrac { b^{ 2 } }{ a } )$ and $L'(ae,\cfrac { -b^{ 2 } }{ a } )$
equation of normal at $L$ and $L'$
at $L$, $\Rightarrow \cfrac { a^{ 2 }x }{ ae } -\cfrac { b^{ 2 }y }{ \cfrac { b^{ 2 } }{ a }  } =a^{ 2 }-b^{ 2 }$
$ \Rightarrow ax-aey=e(a^{ 2 }-b^{ 2 })$
$ \Rightarrow x-ey=\cfrac { a^{ 2 } }{ a } (e^{ 2 })(e)$
$ \Rightarrow x-ey-ae^{ 3 }=0---(i)$
at $L'$ $\Rightarrow \cfrac { a^{ 2 }x }{ ae } -\cfrac { b^{ 2 }y }{ -\cfrac { b^{ 2 } }{ a }  } =a^{ 2 }e^{ 2 }$
$\Rightarrow x+ey-ae^{ 3 }=0---(ii)$

If line $y+3x=c$ is normal of the ellipse ${ x }^{ 2 }+3{ y }^{ 2 }=3$ then equation of normal is-

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $y-3x\pm \sqrt { 3 } =0$

  2. $y+3x\pm \sqrt { 3 } =0$

  3. $y+3x\pm 3 =0$

  4. $y+3x\pm 1 =0$

Show answer
Correct Option: B
Explanation:

If $y=mx+c$ is normal to ellipse
${ c }^{ 2 }={ m }^{ 2 }\cfrac { { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }{ m }^{ 2 } } $
${ x }^{ 2 }+3{ y }^{ 2 }=3$
$\cfrac { { x }^{ 2 } }{ 3 } +\cfrac { { y }^{ 2 } }{ 12 } =1$
${ a }^{ 2 }=3,b=1$
$y+3x=c$
$m=-3$
${ c }^{ 2 }={ (-3) }^{ 2 }\cfrac { { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }{ m }^{ 2 } } =9\times \cfrac { { (3-1) }^{ 2 } }{ 3+9 } $
$=\cfrac { 9\times 4 }{ 12 } =3$

Equation of normal
$y+3x\pm \sqrt { 3 } =0$