The tangent to the parabola $y^2=4x$ at the point where it intersects the circle $x^2+y^2=5$ in the first quadrant, passes through the point:

• A

  $\left(\frac{1}{4},\frac{3}{4}\right)$
• B

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

  $\left(-\frac{1}{4},\frac{1}{2}\right)$
• Correct answer
  D

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

Solution

Let’s first find points of intersection of two curves.
$x^2+4x=5$
$\Rightarrow x=1,-5$ $($ neglect $x=-5$ )
Hence point of intersections are $\left(1,2\right)$ and $\left(1,-2\right).$
Tangent to the parabola at first quadrant point $\left(1,2\right)$
$y\cdot 2=2\left(x+1\right)$
$\Rightarrow y=x+1$ clearly point $\left(\frac{3}{4},\frac{7}{4}\right)$ satisfy the given equation of tangent.

Hence, the correct answer is option: 'D'.