Question

If the parabola ${y^2} = 4x$ meets a circle with centre at $\left( {6,5} \right)$ orthogonally, then possible point(s) of intersection can be
A. $\left( {4,4} \right)$
B. $\left( {9,4} \right)$
C. $\left( {2,8} \right)$
D. $\left( {3,2} \right)$

Answer 1

To solve the question, we will first consider a point of intersection of the parabola and the circle. Then we will find the slope of tangent at that point to the parabola, by differentiating the parabolic curve. Then we will use the concept that if two lines are orthogonal then the product of their slopes is equal to minus one, to find the slope of tangent to the circle at that same point. Then we will find the slope of the tangent to the circle at the point of intersection using the slope of the tangent and also using the centre of the circle and the point of intersection. Then we will equate the two slopes and will find the coordinates of the point of intersection.

Complete step by step answer:
Let the point of intersection of the parabola and the circle be $\left( {{t^2},2t} \right)$. Now, we will find the slope of the tangent to the parabola at this point by differentiating the equation of parabola. So, we have;
${y^2} = 4x$
Differentiating both sides we get;
$\Rightarrow \dfrac{{d{y^2}}}{{dx}} = \dfrac{d}{{dx}}4x$
On simplification we get;
$\Rightarrow 2y\dfrac{{dy}}{{dx}} = 4$

On shifting we get;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{y}$
So, slope of tangent to parabola at $\left( {{t^2},2t} \right)$ is;
$\Rightarrow {\left( {\;\dfrac{{dy}}{{dx}}} \right)_{\left( {{t^2},2t} \right)}} = \dfrac{2}{{2t}}$
On dividing we have;
$\Rightarrow {\left( {\;\dfrac{{dy}}{{dx}}} \right)_{\left( {{t^2},2t} \right)}} = \dfrac{1}{t}$
Let $m$be the slope of the tangent to the circle at $\left( {{t^2},2t} \right)$. Now since the circle and the parabola cut each other orthogonally therefore;
$m \times \dfrac{1}{t} = - 1$
$\therefore m = - t$

Now tangent and normal are perpendicular to each other so, the product of their slopes should be $- 1$. According to this we get the slope of the normal to the circle at point $\left( {{t^2},2t} \right)$ as $\dfrac{1}{t}$.
Now, the centre of the circle is known and one point on the circle is known so we can find the slope of the normal at that point on the circle. So, we have;
Slope of normal $= \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
Putting the point as $\left( {{t^2},2t} \right)$ and $\left( {6,5} \right)$ we get;
Slope of normal $= \dfrac{{2t - 5}}{{{t^2} - 6}}$

Now we have two slopes for the normal. So, we will equate them. So, we get;
$\Rightarrow \dfrac{{2t - 5}}{{{t^2} - 6}} = \dfrac{1}{t}$
On cross multiplication we get;
$\Rightarrow t\left( {2t - 5} \right) = {t^2} - 6$
On solving we get;
$\Rightarrow 2{t^2} - 5t = {t^2} - 6$
On shifting the terms to LHS we get;
$\Rightarrow {t^2} - 5t + 6 = 0$
Splitting the middle term, we get;
$\Rightarrow {t^2} - 3t - 2t + 6 = 0$
$\Rightarrow t\left( {t - 3} \right) - 2\left( {t - 3} \right) = 0$

On further taking common we get;
$\Rightarrow \left( {t - 2} \right)\left( {t - 3} \right) = 0$
From here by equating each term to zero we get $\Rightarrow t = 2,3$.
Now we will put the value in the point of intersection. So, we have;
$\Rightarrow \left( {{t^2},2t} \right) = \left( {{2^2},2 \times 2} \right)$
$\Rightarrow \left( {{t^2},2t} \right) = \left( {4,4} \right)$
Also, by putting $t = 3$ we get;
$\Rightarrow \left( {{t^2},2t} \right) = \left( {{3^2},2 \times 3} \right)$
$\therefore \left( {{t^2},2t} \right) = \left( {9,6} \right)$

Hence, the option A is correct.

Note: To solve these types of questions we need to keep in mind two things. First, we should know what orthogonal intersecting means and how we can use it mathematically. Secondly, we should know that the product of slope of two perpendicular lines is minus one.