# The tangent to the parabola

Question:

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 :

1. (1) $\left(-\frac{1}{3}, \frac{4}{3}\right)$

2. (2) $\left(\frac{1}{4}, \frac{3}{4}\right)$

3. (3) $\left(\frac{3}{4}, \frac{7}{4}\right)$

4. (4) $\left(-\frac{1}{4}, \frac{1}{2}\right)$

Correct Option: , 3

Solution:

To find intersection point of $x^{2}+y^{2}=5$ and $y^{2}=4 x$,

substitute $y^{2}=4 x$ in $x^{2}+y^{2}=5$, we get

$x^{2}+4 x-5=0 \Rightarrow x^{2}+5 x-x-5=0$

$\Rightarrow x(x+5)-1(x+5)=0$

$\therefore x=1,-5$

Intersection point in $1^{\text {st }}$ quadrant be $(1,2)$.

Now, equation of tangent to $y^{2}=4 x$ at $(1,2)$ is

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

$\Rightarrow x-y=0$ .....$\ldots(\mathrm{i})$

Hence, $\left(\frac{3}{4}, \frac{7}{4}\right)$ lies on (i)