Normal at a variable point P on an ellipse frac{{{x^2}}}{{{a^2}}} + frac{{{y^2}}}{{{b^2}}}= 1

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10-2. Parabola, Ellipse, Hyperbola

normal

The normal at a variable point $P$ on an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}= 1$ of eccentricity e meets the axes of the ellipse in $ Q$ and $R$ then the locus of the mid-point of $QR$ is a conic with an eccentricity $e' $ such that :

$e'$ is independent of $e$

$e ' = 1$

$e' = e$

$e' = 1/e$

Solution

Normal at point $P(a \cos \theta, b \sin \theta)$ is

$\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^{2}-b^{2}$

It meets axes at $Q\left(\frac{\left(a^{2}-b^{2}\right) \cos \theta}{a}, 0\right)$

and $R\left(0,-\frac{\left(a^{2}-b^{2}\right) \sin \theta}{b}\right)$

Let $T(h, k)$ is a midpoint of $Q R$ Then $2 h=\frac{\left(a^{2}-b^{2}\right) \cos \theta}{a}$

and $2 k=-\frac{\left(a^{2}-b^{2}\right) \sin \theta}{b}$

$\Rightarrow \cos ^{2} \theta+\sin ^{2} \theta=\frac{4 h^{2} a^{2}}{\left(a^{2}-b^{2}\right)^{2}}+\frac{4 k^{2} b^{2}}{\left(a^{2}-b^{2}\right)^{2}}=1$

$\Rightarrow$ Locus is $\frac{x^{2}}{\frac{\left(a^{2}-b^{2}\right)^{2}}{4 a^{2}}}+\frac{y^{2}}{\frac{\left(a^{2}-b^{2}\right)^{2}}{4 b^{2}}}=1$

which is an ellipse, having eccentricity $e^{\prime},$ given by $e^{r 2}=1-\frac{\frac{\left(a^{2}-b^{2}\right)^{2}}{4 a^{2}}}{\frac{\left(a^{2}-b^{2}\right)^{2}}{4 b^{2}}}=1-\frac{b^{2}}{a^{2}}=e^{2}$

$e^{\prime}=e$

Note : In Equation (ii), $\frac{\left(a^{2}-b^{2}\right)}{4 a^{2}}<\frac{\left(a^{2}-b^{2}\right)}{4 b^{2}}$. Hence, $x$ -axis is minor axis.

Standard 11

Mathematics

Find the equation of the ellipse, with major axis along the $x-$ axis and passing through the points $(4,\,3)$ and $(-1,\,4)$

medium

The locus of a variable point whose distance from $(-2, 0)$ is $\frac{2}{3}$ times its distance from the line $x = – \frac{9}{2}$, is

medium

(IIT-1994)

Let $\mathrm{A}(\alpha, 0)$ and $\mathrm{B}(0, \beta)$ be the points on the line $5 x+7 y=50$. Let the point $P$ divide the line segment $A B$ internally in the ratio $7: 3$. Let $3 x-$ $25=0$ be a directrix of the ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the corresponding focus be $S$. If from $S$, the perpendicular on the $\mathrm{x}$-axis passes through $\mathrm{P}$, then the length of the latus rectum of $\mathrm{E}$ is equal to

hard

(JEE MAIN-2024)

Let $F_1\left(x_1, 0\right)$ and $F_2\left(x_2, 0\right)$, for $x_1<0$ and $x_2>0$, be the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{8}=1$. Suppose a parabola having vertex at the origin and focus at $F_2$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant.

($1$)The orthocentre of the triangle $F_1 M N$ is

($A$) $\left(-\frac{9}{10}, 0\right)$ ($B$) $\left(\frac{2}{3}, 0\right)$ ($C$) $\left(\frac{9}{10}, 0\right)$ ($D$) $\left(\frac{2}{3}, \sqrt{6}\right)$

($2$) If the tangents to the ellipse at $M$ and $N$ meet at $R$ and the normal to the parabola at $M$ meets the $x$-axis at $Q$, then the ratio of area of the triangle $M Q R$ to area of the quadrilateral $M F_{\mathrm{I}} N F_2$ is

($A$) $3: 4$ ($B$) $4: 5$ ($C$) $5: 8$ ($D$) $2: 3$

Givan the answer qestion ($1$) and ($2$)

normal

(IIT-2016)

Let $\theta$ be the acute angle between the tangents to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ and the circle $x^{2}+y^{2}=3$ at their point of intersection in the first quadrant. Then $\tan \theta$ is equal to :

hard

(JEE MAIN-2021)

The normal at a variable point $P$ on an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}= 1$ of eccentricity e meets the axes of the ellipse in $ Q$ and $R$ then the locus of the mid-point of $QR$ is a conic with an eccentricity $e' $ such that :

Solution

Similar Questions

Find the equation of the ellipse, with major axis along the $x-$ axis and passing through the points $(4,\,3)$ and $(-1,\,4)$

The locus of a variable point whose distance from $(-2, 0)$ is $\frac{2}{3}$ times its distance from the line $x = – \frac{9}{2}$, is

Let $\theta$ be the acute angle between the tangents to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ and the circle $x^{2}+y^{2}=3$ at their point of intersection in the first quadrant. Then $\tan \theta$ is equal to :

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