Question

Three tangents to a parabola, which are such that the tangents of their inclinations to the axis are in a given harmonic progression, form a triangle is
(a) which is isosceles
(b) equilateral
(c) having constant area
(d) None of these

Answer 1

Hint: Equations of tangents to the parabola ${{y}^{2}}=4ax$ at point $P(a{{t}_{1}}^{2},2a{{t}_{1}})$ , $Q(a{{t}_{2}}^{2},2a{{t}_{2}})$ and $R(a{{t}_{3}}^{2},2a{{t}_{3}})$ is given by ${{t}_{1}}y=x+a{{t}_{1}}^{2}$ , ${{t}_{2}}y=x+a{{t}_{2}}^{2}$ and ${{t}_{3}}y=x+a{{t}_{3}}^{2}$ having slope $\dfrac{1}{{{t}_{1}}}$ , $\dfrac{1}{{{t}_{2}}}$ and $\dfrac{1}{{{t}_{3}}}$ respectively. Then ${{t}_{1}},{{t}_{2}},{{t}_{3}}$ are in arithmetic progression (A.P) . Find the area of triangle using formula $\dfrac{1}{2}\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|$ if $({{x}_{1}},{{y}_{1}})$ , $({{x}_{2}},{{y}_{2}})$ and $({{x}_{3}},{{y}_{3}})$ be three points.

Complete step by step answer:
Let us consider a parabola ${{y}^{2}}=4ax$ , then take three points on the parabola $P(a{{t}_{1}}^{2},2a{{t}_{1}})$ , $Q(a{{t}_{2}}^{2},2a{{t}_{2}})$ and $R(a{{t}_{3}}^{2},2a{{t}_{3}})$ . Now equations of tangents to the parabola ${{y}^{2}}=4ax$ at point $P,Q,R$ is given by ${{t}_{1}}y=x+a{{t}_{1}}^{2}$ , ${{t}_{2}}y=x+a{{t}_{2}}^{2}$ and ${{t}_{3}}y=x+a{{t}_{3}}^{2}$ . These tangents have slope $\dfrac{1}{{{t}_{1}}}$ , $\dfrac{1}{{{t}_{2}}}$ and $\dfrac{1}{{{t}_{3}}}$ respectively. It is given in the question that these slopes are in harmonic progression (H.P).
Therefore ${{t}_{1}},{{t}_{2}},{{t}_{3}}$are in arithmetic progression (A.P) because a series is said to be in H.P when the reciprocals of the terms form an A.P . Let $d$ be the common difference of given A.P ${{t}_{1}},{{t}_{2}},{{t}_{3}}$ then,
${{t}_{2}}-{{t}_{1}}=d$
${{t}_{3}}-{{t}_{2}}=d$
And ${{t}_{3}}-{{t}_{1}}=2d$
Now we will find the intersection points of tangent drawn from point $Q$ and $R$ , $R$ and $P$ , $P$ and $R$ . Let the point of intersection of tangents be A, B and C respectively. Then the coordinates of A, B and C are
$A=\{a{{t}_{2}}{{t}_{3}},a({{t}_{2}}+{{t}_{3}})\}$
$B=\{a{{t}_{3}}{{t}_{1}},a({{t}_{3}}+{{t}_{1}})\}$
$C=\{a{{t}_{1}}{{t}_{2}},a({{t}_{1}}+{{t}_{2}})\}$
We will now find the area of triangle formed by points A, B and C. If $({{x}_{1}},{{y}_{1}})$ , $({{x}_{2}},{{y}_{2}})$ and $({{x}_{3}},{{y}_{3}})$ be three points then area of triangle is given by the formula $\dfrac{1}{2}\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|$ Using this formula for area of triangle and coordinate points of A, B and C we will find the area of triangle $ABC$ we get,
Area of $\vartriangle ABC=\dfrac{1}{2}\left| \begin{matrix}
   a{{t}_{2}}{{t}_{3}} & a({{t}_{2}}+{{t}_{3}}) & 1 \\
   a{{t}_{3}}{{t}_{1}} & a({{t}_{3}}+{{t}_{1}}) & 1 \\
   a{{t}_{1}}{{t}_{2}} & a({{t}_{1}}+{{t}_{2}}) & 1 \\
\end{matrix} \right|$
Now we will find the value of determinant by using row and column transformation. For solving the determinant we will take out $a$ from column 1 and column 2.
$=\dfrac{1}{2}{{a}^{2}}\left| \begin{matrix}
   {{t}_{2}}{{t}_{3}} & ({{t}_{2}}+{{t}_{3}}) & 1 \\
   {{t}_{3}}{{t}_{1}} & ({{t}_{3}}+{{t}_{1}}) & 1 \\
   {{t}_{1}}{{t}_{2}} & ({{t}_{1}}+{{t}_{2}}) & 1 \\
\end{matrix} \right|$
Subtract row 2 and row 1 and substitute in row 2. Similarly subtract row 3 and row 1 and substitute in row 3 we get,
$=\dfrac{1}{2}{{a}^{2}}\left| \begin{matrix}
   {{t}_{2}}{{t}_{3}} & ({{t}_{2}}+{{t}_{3}}) & 1 \\
   {{t}_{3}}({{t}_{1}}-{{t}_{2}}) & ({{t}_{1}}-{{t}_{2}}) & 0 \\
   {{t}_{2}}({{t}_{1}}-{{t}_{3}}) & ({{t}_{1}}-{{t}_{3}}) & 0 \\
\end{matrix} \right|$
Taking $({{t}_{1}}-{{t}_{3}})$ and $({{t}_{1}}-{{t}_{2}})$ common from row 3 and row 2 respectively we get,
$=\dfrac{1}{2}{{a}^{2}}({{t}_{1}}-{{t}_{2}})({{t}_{1}}-{{t}_{3}})\left| \begin{matrix}
   {{t}_{2}}{{t}_{3}} & ({{t}_{2}}+{{t}_{3}}) & 1 \\
   {{t}_{3}} & 1 & 0 \\
   {{t}_{2}} & 1 & 0 \\
\end{matrix} \right|$
Now solving the determinant we get,
$=\dfrac{1}{2}{{a}^{2}}\left| ({{t}_{1}}-{{t}_{2}})({{t}_{1}}-{{t}_{3}})({{t}_{2}}-{{t}_{3}}) \right|$
Earlier we have found out the value of ${{t}_{2}}-{{t}_{1}}=d$ , ${{t}_{3}}-{{t}_{2}}=d$ and ${{t}_{3}}-{{t}_{1}}=2d$ . Substituting these values above we get,
$\begin{align}
  & =\dfrac{1}{2}{{a}^{2}}(d)(d)(2d) \\
 & ={{a}^{2}}{{d}^{3}} \\
\end{align}$
Since the area of the triangle has come out to be a constant value so three tangents to a parabola, which are such that the tangents of their inclinations to the axis are in a given harmonic progression, form a triangle is having constant area. Hence, option (c) is correct.

Note:
If $({{x}_{1}},{{y}_{1}})$ , $({{x}_{2}},{{y}_{2}})$ and $({{x}_{3}},{{y}_{3}})$ be three points then area of triangle is given by the formula $\dfrac{1}{2}\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|$ Using this formula for the area of triangle and coordinate points of A, B and C we will find the area of triangle ABC . Area of the triangle is a positive quantity so care should be taken when we calculate, if the value of the determinant turns out to be negative then take the modulus value of the result.