Question

Area of a triangle formed by tangent and normal to the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $P(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}})$ with the X-axis is

• A. $4ab$
• B. $\frac{ab\sqrt{a^2+b^2}}{4}$
• C. $\frac{ab\sqrt{a^2-b^2}}{4}$
• D. $\frac{b(a^2+b^2)}{4a}$

Answer 1

Answer: (D)

Given curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and point $P\left(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right)$
Slope of tangent at $P$ is $-b/a$
Slope of normal at $P$ is $a/b$.
Equation of tangent on the curve
$bx+ay=\sqrt{2}ab$...(i)
Equation of normal on the tangent of the curve
$-ax+by=\frac{\left(b^2-a^2\right)}{\sqrt{2}}$...(ii)
Equation of $x$ -axis $y=0$...(iii)
Solving these equations in successive manner, we get three points of triangle
$\left(\frac{a^2-b^2}{a\sqrt{2}},0\right),(\sqrt{2}\cdot a,0)$ and $\left(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right)$
Now, area of triangle $=\frac{1}{2}\left|\begin{array}{ccc}\frac{a^2-b^2}{a\sqrt{2}}& 0& 1\\ \sqrt{2}\cdot a& 0& 1\\ a/\sqrt{2}& b/\sqrt{2}& 1\end{array}\right|$
$=\frac{-b}{2\sqrt{2}}\left\{\frac{a^2-b^2}{a\sqrt{2}}-\sqrt{2}\cdot a\right\}=\frac{-b}{4a}\left(-a^2-b^2\right)$
$=\frac{b\left(a^2+b^2\right)}{4a}$