SATHEE: Area Question 7

Area Question 7

Question 7

The tangent to the parabola $y^{2} = 4 x$ at the point where it intersects the circle $x^{2} + y^{2} = 5$ in the first quadrant, passes through the point (a) $\frac{1}{4}, \frac{3}{4}$ (b) $\frac{3}{4}, \frac{7}{4}$ (c) $- \frac{1}{3}, \frac{4}{3}$ (d) $- \frac{1}{4}, \frac{1}{2}$

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Solution:

Given equations of the parabola $y^{2} = 4 x$ and circle

$x^{2} + y^{2} = 5$

So, for point of intersection of curves (i) and (ii), put $y^{2} = 4 x$ in Eq. (ii), we get

$\begin{array}{lc} \Rightarrow & x^{2} + 4 x - 5 = 0 \Rightarrow & (x - 1) (x + 5) = 0 \Rightarrow & x = 1, - 5 \end{array}$

For first quadrant $x = 1$ , so $y = 2$ .

Now, equation of tangent of parabola (i) at point $(1, 2)$ is $T = 0$

$\Rightarrow 2 y = 2 (x + 1)$

$\Rightarrow x - y + 1 = 0$

The point $\frac{3}{4}, \frac{7}{4}$ satisfies, the equation of line

$x - y + 1 = 0$

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