Question

Intersection point of the tangents drawn to the parabola $ {{y}^{2}}=4ax $ from the points $ (at_{1}^{2},2a{{t}_{1}}) $ and $ (at_{2}^{2},2a{{t}_{2}}) $ is

• A. $ (a{{t}_{1}}{{t}_{2}},a({{t}_{1}}+{{t}_{2}})) $
• B. $ (a({{t}_{1}}+{{t}_{2}}),a{{t}_{1}}{{t}_{2}}) $
• C. $ (t_{1}^{2}t_{2}^{2},{{t}_{1}}+{{t}_{2}}) $
• D. None of these

Answer 1

Answer: (A)

Equation of tangent to the parabola
$ {{y}^{2}}=4ax $ at the point $ (at_{1}^{2},2a{{t}_{1}}) $
is $ 2a{{t}_{1}}=2a(x+at_{1}^{2}) $
$ \Rightarrow $ $ y{{t}_{1}}=x+at_{1}^{2} $ ...(i)
Similarly, equation of tangent at the point
$ (at_{2}^{2},2a{{t}_{2}}) $ is $ 2ya{{t}_{2}}=2a(x+at_{2}^{2}) $
$ \Rightarrow $ $ y{{t}_{2}}=x+at_{2}^{2} $ ...(ii)
On solving Eqs. (i) and (ii), we get
$ y=a({{t}_{1}}+{{t}_{2}}),x=a{{t}_{1}}{{t}_{2}} $
$ \therefore $ Intersection point $ [a{{t}_{1}}{{t}_{2}},a({{t}_{1}}+{{t}_{2}})] $