Question

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) $\left( \dfrac{-1}{3},\dfrac{4}{3} \right)$
(b) $\left( \dfrac{-1}{4},\dfrac{1}{2} \right)$
(c) $\left( \dfrac{3}{4},\dfrac{7}{4} \right)$
(d) $\left( \dfrac{1}{4},\dfrac{3}{4} \right)$

Answer 1

Hint: First we will find the point of intersection of the parabola ${{y}^{2}}=4x$ and the circle ${{x}^{2}}+{{y}^{2}}=5$. Then we will find the equation of the tangent to the parabola by the formula $y{{y}_{1}}=2a\left( x+{{x}_{1}} \right)$. Once we have the equation, we will substitute the points mentioned in the question to see which point satisfies the equation of tangent.

Complete step-by-step answer:
To find the point of intersection of the parabola ${{y}^{2}}=4x\text{ }\ldots \left( i \right)$
And the circle ${{x}^{2}}+{{y}^{2}}=5\text{ }\left( ii \right)$
We will substitute the value of equation (i) in equation (ii). We get
$\begin{align}
  & {{x}^{2}}+4x=5 \\
 & \Rightarrow {{x}^{2}}+4x-5=0 \\
\end{align}$
Solving this equation further, we get
$\begin{align}
  & {{x}^{2}}-5x+x-5=0 \\
 & \Rightarrow x\left( x-5 \right)+\left( x-5 \right)=0 \\
 & \Rightarrow \left( x-5 \right)\left( x+1 \right)=0 \\
\end{align}$
Hence, the value turns out to be 1 and -5.

But the question says that the tangent to the parabola and a point of the circle intersects in the first quadrant. Thus, the negative value of x, that is, -5 can not be used. We will consider the value of x as 1.

When $x=1$, the value of y can be found by putting it in equation (i).
Thus, we get
$\begin{align}
  & {{y}^{2}}=4\cdot 1=4 \\
 & \Rightarrow y=\pm 2 \\
\end{align}$
Again, y can not be negative as the point is in the first quadrant.
Therefore, the point of intersection is $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 1,2 \right)$.
The equation of the parabola is ${{y}^{2}}=4x$ which when compared to the general equation of the parabola ${{y}^{2}}=4ax$ gives $a=1$.

We know that the equation of the tangent to the parabola is given by
$y{{y}_{1}}=2a\left( x+{{x}_{1}} \right)$

Substituting the values of a and $\left( {{x}_{1}},{{y}_{1}} \right)$in the above equation, we get
 $\begin{align}
  & 2y=2\cdot 1\left( x+1 \right) \\
 & 2y=2x+2 \\
 & y=x+1
\end{align}$

This can be rearranged as $x-y+1=0$.
Clearly, out of all the points mentioned in the question, point $\left( \dfrac{3}{4},\dfrac{7}{4} \right)$ satisfies the above equation of tangent.

So, the correct answer is “Option C”.

Note: To verify our answer, we should always consider other options. After putting the points $\left( \dfrac{-1}{3},\dfrac{4}{3} \right)$ , $\left( \dfrac{-1}{4},\dfrac{1}{2} \right)$ , and $\left( \dfrac{1}{4},\dfrac{3}{4} \right)$ in the equation $x-y+1=0$, we can see that they do not satisfy the equation.