Question

The tangent at the point $P\left( {{x}_{1}},{{y}_{1}} \right)$to the parabola${{y}^{2}}=4ax$meets the parabola${{y}^{2}}=4a\left( x+b \right)$at Q and R, the coordinates of the midpoint of QR are
A) $\left( {{x}_{1}}-a,{{y}_{1}}+b \right)$
B) $\left( {{x}_{1}},{{y}_{1}} \right)$
C) $\left( {{x}_{1}}+b,{{y}_{1}}+a \right)$
D) $\left( {{x}_{1}}-b,{{y}_{1}}-b \right)$

Answer 1

Here we have to find the midpoint of QR. For that, we will first put the point $P\left( {{x}_{1}},{{y}_{1}} \right)$in the equation of the parabola${{y}^{2}}=4ax$. Then we will find the value of y in terms of. The roots of the equation formed will be${{x}_{2}}\text{ }\\!\\!\And\\!\\!\text{ }{{x}_{3}}.$, from there, we will get the value of x coordinate of midpoint of QR. Then we will find the value of x in terms of${{x}_{1}}\And {{y}_{1}}$. The roots of the equation formed will be${{y}_{2}}\text{ }\\!\\!\And\\!\\!\text{ }{{\text{y}}_{3}}.$, from there, we will get the value of y coordinate of midpoint of QR.

Complete step by step solution:
It is given that the tangent at the point $P\left( {{x}_{1}},{{y}_{1}} \right)$ to the parabola ${{y}^{2}}=4ax$meets the parabola at R and Q.
Equation of the tangent at the point$P\left( {{x}_{1}},{{y}_{1}} \right)$ to the parabola ${{y}^{2}}=4ax$is given by
$y{{y}_{1}}=2a\left( x+{{x}_{1}} \right)$ ………….$\left( 1 \right)$
As it is also given that the above line intersects the parabola at Q and R.
Let the coordinate of point Q and point R be $\left( {{x}_{2}},{{y}_{2}} \right)\And \left( {{x}_{3}},{{y}_{3}} \right)$
Now, we will put the value of y from equation 1 in the equation of parabola${{y}^{2}}=4a\left( x+b \right)$
$~{{\left[ \dfrac{2a\left( x+{{x}_{1}} \right)}{{{y}_{1}}} \right]}^{2}}=\text{ }4a\left( x+b \right)$
On simplifying, we get
$ax{}^\text{2}\text{ }+\text{ }x\left( 2a{{x}_{1}}-{{y}_{1}}{}^\text{2} \right)\text{ }+\text{ }a{{x}_{1}}{}^\text{2}-{{y}_{1}}{}^\text{2}b\text{ }=\text{ }0$
For the above equation, roots are ${{x}_{2}}\text{ }\\!\\!\And\\!\\!\text{ }{{x}_{3}}.$
=> ${{x}_{2}}+\text{ }{{x}_{3}}=\text{ }\dfrac{({{y}_{1}}{}^\text{2}\text{ }-\text{ }2a{{x}_{1}})}{a}$ ……………$\left( 2 \right)$
But$P\left( {{x}_{1}},{{y}_{1}} \right)$ lies on the parabola${{y}^{2}}=4ax$, so it will satisfy the equation of parabola.
Therefore,
${{y}_{1}}^{2}=4a{{x}_{1}}$
Now, we will put the value of ${{y}_{1}}^{2}$in equation 2.
${{x}_{2}}+\text{ }{{x}_{3}}\text{=}\dfrac{\left( 4a{{x}_{1}}-\text{ }2a{{x}_{1}} \right)}{a}\text{ }=\dfrac{2a{{x}_{1}}}{a}=2{{x}_{1}}$
Thus,
${{x}_{1}}=\dfrac{{{x}_{2}}+\text{ }{{x}_{3}}}{2}$
Now, we will put the value of x from equation 1 in the equation of parabola${{y}^{2}}=4a\left( x+b \right)$
${{y}^{2}}=4a\left( \dfrac{y{{y}_{1}}}{2a}-{{x}_{1}}+b \right)$
On simplifying, we get
$y{}^\text{2}-2\text{y}{{\text{y}}_{1}}+4a{{x}_{1}}-4ab\text{ }=\text{ }0$
For the above equation, roots are ${{y}_{2}}\text{ }\\!\\!\And\\!\\!\text{ }{{\text{y}}_{3}}.$
[\begin{aligned}
& {{y}{2}}+\text{ }{{\text{y}}{3}}=\text{ }\dfrac{2{{y}{1}}}{1} \\
& \dfrac{{{y}{2}}+\text{ }{{\text{y}}{3}}}{2}={{y}{1}} \\
\end{aligned}] ……………$\left( 3 \right)$
Hence, the coordinates of the midpoint of the chord QR are $\left( {{x}_{1}},{{y}_{1}} \right)$

Thus, the correct answer is option B.

Note:
Remember that the point which lies on the given curve or line will satisfy the equation of the same curve or line i.e. the value of coordinates of the point will satisfy the equation of curve or line.
Intersection of a line and a curve means there is a common point which lies both on the line and the curve and satisfies the equation of both curve and line.