The area (in sq. units) of the triangle forme | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the point be $P = \left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$. The equation of the tangent at $P$ to the ellipse $\frac{x^2}{a^2}+\fr

Vedclass · Accepted Answer

$\frac{b}{4 a}\left(a^2+b^2\right)$

Vedclass · Answer

(C) Let the point be $P = \left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}ight)$. The equation of the tangent at $P$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{x}{a\sqrt{2}} + \frac{y}{b\sqrt{2}} = 1$. The $X$-intercept $M$ is found by setting $y=0$,giving $x = a\sqrt{2}$,so $M = (a\sqrt{2}, 0)$.The slope of the tangent is $m_t = -\frac{b}{a}$. The slope of the normal is $m_n = \frac{a}{b}$. The equation of the normal at $P$ is $y - \frac{b}{\sqrt{2}} = \frac{a}{b} \left(x - \frac{a}{\sqrt{2}}ight)$. The $X$-intercept $N$ is found by setting $y=0$,giving $-\frac{b}{\sqrt{2}} = \frac{a}{b} \left(x - \frac{a}{\sqrt{2}}ight)$,which simplifies to $x = \frac{a^2-b^2}{a\sqrt{2}}$.The base of the triangle $MN$ is $|x_M - x_N| = |a\sqrt{2} - \frac{a^2-b^2}{a\sqrt{2}}| = |\frac{2a^2 - a^2 + b^2}{a\sqrt{2}}| = \frac{a^2+b^2}{a\sqrt{2}}$.The height of the triangle is the $y$-coordinate of $P$,which is $\frac{b}{\sqrt{2}}$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{a^2+b^2}{a\sqrt{2}} 	imes \frac{b}{\sqrt{2}} = \frac{b(a^2+b^2)}{4a}$.

The area (in sq. units) of the triangle formed by the tangent and normal at a point $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$ to the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the $X$-axis is

Similar Questions

The equation of the ellipse whose axes are the coordinate axes,which passes through the point $(-3, 1)$,and has an eccentricity $e = \sqrt{\frac{2}{5}}$ is:

The locus of mid-points of the line segments joining $(-3,-5)$ and the points on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ is :

Let the equation of an ellipse be $\frac{x^{2}}{144}+\frac{y^{2}}{25}=1$. Then, the radius of the circle with center $(0, \sqrt{2})$ and passing through the foci of the ellipse is

Find the position of the point $(4, -3)$ with respect to the ellipse $2x^2 + 5y^2 = 20$.

The eccentric angle of the point $(2, \sqrt{3})$ lying on $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module