Question

Locus of the intersection of the tangents at the ends of the normal chords of the parabola \[{{y}^{2}}=4ax\] is
(a) \[\left( 2a+x \right){{y}^{2}}+4{{a}^{3}}=0\]
(b) \[\left( 2a-x \right){{y}^{2}}+4{{a}^{3}}=0\]
(c) \[\left( 2a-x \right){{y}^{2}}-4{{a}^{3}}=0\]
(d) None of these

Answer 1

Hint: Write the equation of the tangent to the parabola in parametric form, now use the condition of normal chord intersect again on the parabola, that is the relation between two parametric coordinates. Solve for x and y in terms of parametric coordinate and eliminate at last to get the locus.

Complete step-by-step answer:

We have the parabola \[{{y}^{2}}=4ax\]. We have to find the locus of point of intersection of the tangents at two points on the parabola such that the line joining two points forms the normal of the parabola.
Consider any two points on the parabola \[{{y}^{2}}=4ax\] of the form \[P\left( {{t}_{1}} \right)=\left( at_{1}^{2},2a{{t}_{1}} \right)\] and \[Q\left( {{t}_{2}} \right)=\left( at_{2}^{2},2a{{t}_{2}} \right)\].
We know that the equation of any normal of the parabola \[{{y}^{2}}=4ax\] in parametric form is \[y+tx=2at+a{{t}^{3}}\].
We also know that if this normal intersects the parabola at two points \[\left( {{t}_{1}} \right)\] and \[\left( {{t}_{2}} \right)\], then we have\[{{t}_{2}}=-\left( {{t}_{1}}+\dfrac{2}{{{t}_{1}}} \right).....\left( 1 \right)\].
We now have to write the equation of tangents at the two points.
We know that the equation of tangent to the parabola \[{{y}^{2}}=4ax\] in parametric form is \[ty=x+a{{t}^{2}}\].
Thus, the equation of tangent at \[P\left( {{t}_{1}} \right)\] is \[{{t}_{1}}y=x+at_{1}^{2}.....\left( 2 \right)\].
The equation of tangent at \[Q\left( {{t}_{2}} \right)\] is \[{{t}_{2}}y=x+at_{2}^{2}.....\left( 3 \right)\].
By substituting equation (2) in equation (3), we get \[-\left( {{t}_{1}}+\dfrac{2}{{{t}_{1}}} \right)y=x+a{{\left( {{t}_{1}}+\dfrac{2}{{{t}_{1}}} \right)}^{2}}\].
Solving the above equation, we have \[-{{t}_{1}}y-\dfrac{2y}{{{t}_{1}}}=x+at_{1}^{2}+\dfrac{4a}{t_{1}^{2}}+4a.....\left( 4 \right)\].
We have to find the point of intersection of two tangents. We will subtract equation (4) from equation (2) and obtain \[2{{t}_{1}}y+\dfrac{2y}{{{t}_{1}}}=\dfrac{4a}{t_{1}^{2}}+4a\].
Solving the above equation by taking LCM, we have \[\dfrac{2y\left( t_{1}^{2}+1 \right)}{{{t}_{1}}}=\dfrac{4a\left( 1+t_{1}^{2} \right)}{t_{1}^{2}}\].
\[\Rightarrow y=\dfrac{2a}{{{t}_{1}}}\]
Substituting the above value in equation (2), we have \[{{t}_{1}}\left( \dfrac{2a}{{{t}_{1}}} \right)=x+at_{1}^{2}\].
\[\Rightarrow x=2a-at_{1}^{2}\]
Thus, we have \[x=2a-at_{1}^{2},y=\dfrac{2a}{{{t}_{1}}}\] as the point of intersection of two curves.
Eliminating the variable \[{{t}_{1}}\] by rearranging the terms, we have \[x=2a-a{{\left( \dfrac{2a}{y} \right)}^{2}}\].
\[\Rightarrow \left( 2a-x \right){{y}^{2}}=4{{a}^{3}}\]
Hence, the locus of the intersection of the tangents at the ends of the normal chords of the parabola \[{{y}^{2}}=4ax\] is \[\left( 2a-x \right){{y}^{2}}-4{{a}^{3}}=0\], which is option (c).

Note: We can also solve this question by taking the equation of normal of the parabola in slope form and then finding the points on the parabola at which the normal intersects the parabola. We can write an equation of tangents at those points and then equate them to find their point of intersection.