Question

Locus of the point of intersection of the normals at the ends of parallel chords of gradient $m$ of the parabola ${{y}^{2}}=4ax$ is
(a). $2{{m}^{2}}x-{{m}^{3}}y=4a(2+{{m}^{2}})$
(b). $2{{m}^{2}}x+{{m}^{3}}y=4a(2+{{m}^{2}})$
(c). $2mx+{{m}^{2}}y=4a(2+m)$
(d). $2{{m}^{2}}x-{{m}^{3}}y=4a(2-{{m}^{2}})$

Answer 1

Hint: To find the locus of point of intersection of the normals at the ends of parallel chords of given slope of the parabola, write the equation of chord joining any two points on the parabola and the equation of normal to the parabola and then compare the two equations to find the locus.

Complete step by step answer:
We have a parabola ${{y}^{2}}=4ax$. We want to find the locus of point of intersection of the normals at the ends of parallel chords of given slope of the parabola.
We know that the equation of chord joining two points $A({{t}_{1}})=(at_{1}^{2},2a{{t}_{1}})$ and $B({{t}_{2}})=(at_{2}^{2},2a{{t}_{2}})$ on the parabola ${{y}^{2}}=4ax$ is $y({{t}_{1}}+{{t}_{2}})=2x+2a{{t}_{1}}{{t}_{2}}$ .
Dividing the equation by ${{t}_{1}}+{{t}_{2}}$ , we get $y=\dfrac{2x}{{{t}_{1}}+{{t}_{2}}}+\dfrac{2a{{t}_{1}}{{t}_{2}}}{{{t}_{1}}+{{t}_{2}}}$
As the slope of chord is $m$ , we get $m=\dfrac{2}{{{t}_{1}}+{{t}_{2}}}$
$\Rightarrow {{t}_{1}}+{{t}_{2}}=\dfrac{2}{m}$ $(1)$
We know that the equation of normal at any point $(a{{t}^{2}},2at)$ of the parabola ${{y}^{2}}=4ax$ is $y+tx=2at+a{{t}^{3}}$
Thus, the equation of normal at $A({{t}_{1}})$ is $y+{{t}_{1}}x=2a{{t}_{1}}+at_{1}^{3}$ .
The equation of normal at $B({{t}_{2}})$ is $y+{{t}_{2}}x=2a{{t}_{2}}+at_{2}^{3}$
We know their point of intersection is $(a(t_{1}^{2}+t_{2}^{2}+{{t}_{1}}{{t}_{2}}+2),-a{{t}_{1}}{{t}_{2}}({{t}_{1}}+{{t}_{2}}))$
Let’s assume that locus of point of intersection is $(x,y)$
Thus, we have
$\Rightarrow x=a(t_{1}^{2}+t_{2}^{2}+{{t}_{1}}{{t}_{2}}+2),y=-a{{t}_{1}}{{t}_{2}}({{t}_{1}}+{{t}_{2}})$ $(2)$
We know ${{({{t}_{1}}+{{t}_{2}})}^{2}}=t_{1}^{2}+t_{2}^{2}+2{{t}_{1}}{{t}_{2}}=\dfrac{4}{{{m}^{2}}}$
$\Rightarrow t_{1}^{2}+t_{2}^{2}=\dfrac{4}{{{m}^{2}}}-2{{t}_{1}}{{t}_{2}}$ $(3)$
Now, substituting the values of equation $(1)$ and $(3)$ in equation $(2)$ , we get $x=a(\dfrac{4}{{{m}^{2}}}-2{{t}_{1}}{{t}_{2}}+{{t}_{1}}{{t}_{2}}+2)$ and $y=-a{{t}_{1}}{{t}_{2}}\dfrac{2}{m}$
To solve the equation involving $x$ , divide both sides by $a$ and rearrange the terms to get equation in terms of ${{t}_{1}}{{t}_{2}}$ . Similarly, to solve the equation involving $y$ , multiply the equation on both sides by $\dfrac{m}{a}$ .
Thus, we get $\dfrac{x}{a}-\dfrac{4}{{{m}^{2}}}-2=-{{t}_{1}}{{t}_{2}}$ and $\dfrac{my}{a}=-2{{t}_{1}}{{t}_{2}}$
Substituting the value from the second equation in the first equation, we get $\dfrac{x}{a}-\dfrac{4}{{{m}^{2}}}-2=\dfrac{my}{2a}$
Rearranging the terms, taking LCM and then cross multiplying the terms, we get

& \Rightarrow 2x-my=2a(\dfrac{4}{{{m}^{2}}}+2) \\\ & \Rightarrow {{m}^{2}}(2x-my)=2a(4+2{{m}^{2}}) \\\ & \Rightarrow 2{{m}^{2}}x-{{m}^{3}}y=4a(2+{{m}^{2}}) \\\ \end{aligned}$$ We observe that this is the equation of a straight line with two variable parameters. However, it is a cubic equation with respect to the variable m. Hence, the correct answer is $$2{{m}^{2}}x-{{m}^{3}}y=4a(2+{{m}^{2}})$$ Note: We can also solve this question by writing the equation of chord of the parabola in slope form and then use it to write the equation of normal of the parabola.