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NDA Math Conic Section

Question : 10 of 34

Marks: +1, -0

y^{2}=x,

x

\theta?

Solution: 👈: Video Solution

Given parabola :

y^{2}=x

From standard equation of parabola

y^{2}=4 a x

Focus

=(a, 0)

Here

a=\frac{1}{4}

∴ Focus of parabola

y^{2}=x=\left(\frac{1}{4}, 0\right)

Vertex of the parabola

=(0,0)

Let

P\left(\frac{t^{2}}{4}, \frac{t}{2}\right)

lies on the parabola then, its slope

=\tan(\theta)

\therefore \frac{\frac{t}{2}}{\frac{t^{2}}{4}}=\tan\theta\Rightarrow t=2\cot\theta

So, point

P\left(\frac{t^{2}}{4}, \frac{t}{2}\right)=(\cot^{2}\theta, \cot\theta)

Its distance from vertex

(0,0)

=\sqrt{(\cot^{2}\theta-0)^{2}+(\cot\theta-0)^{2}}

=\cot\theta\sqrt{\cot^{2}\theta+1}

=\cot\theta \cdot \text{cosec}\theta

=\cos\theta \cdot \text{cosec}^{2}\theta

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