Let the tangent to the parabola

Question:

Let the tangent to the parabola $\mathrm{S}: \mathrm{y}^{2}=2 \mathrm{x}$ at the point $\mathrm{P}(2,2)$ meet the $\mathrm{X}$-axis at $\mathrm{Q}$ and normal at it meet the parabola $S$ at the point $R$. Then the area (in sq. units) of the triangle PQR is equal to:

  1. $\frac{25}{2}$

  2. $\frac{35}{2}$

  3. $\frac{15}{2}$

  4. 25


Correct Option: 1

Solution:

Tangent at $\mathrm{P}: \mathrm{y}(2)=2(1 / 2)(\mathrm{x}+2)$

$\Rightarrow 2 y=x+2$

$\therefore Q=(-2,0)$

Normal at $\mathrm{P}: \mathrm{y}-2=-\frac{(2)}{2 \cdot 1 / 2}(\mathrm{x}-2)$

$\Rightarrow \mathrm{y}-2=-2(\mathrm{x}-2)$

$\Rightarrow \mathrm{y}=6-2 \mathrm{x}$

$\therefore$ Solving with $y^{2}=2 x \Rightarrow R\left(\frac{9}{2}-3\right)$

$\therefore \operatorname{Ar}(\Delta \mathrm{PQR})=\frac{1}{2}\left|\begin{array}{ccc}2 & 2 & 1 \\ -2 & 1 & 1 \\ \frac{9}{2} & 3- & 1\end{array}\right|$

$=\frac{25}{2}$ sq.units

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