Question

The points of contact Q and R of tangent from the point P (2, 3) on the parabola ${{y}^{2}}=4x$ are:
(a)(9, 6) and (1, 2)
(b)(1, 2) and (4, 4)
(c)(4, 4) and (9, 6)
(d)(9, 6) and $\left( \dfrac{1}{4},1 \right)$

Answer 1

Hint: We know that if a parabola is of the form: ${{y}^{2}}=4ax$ and then the tangent at any point on this parabola is equal to $y=mx+\dfrac{a}{m}$. Now, the parabola given in the above problem is ${{y}^{2}}=4x$ and here the value of $a=1$ so substituting this value of $''a''$ in the equation of tangent we will get the equation of a tangent. Then pass the point P (2, 3) in this equation of a tangent. Then we will get a quadratic equation in “m” and then solve it and get the values of m. Now, let us assume the two points of contacts Q and R be $\left( {{t}_{1}}^{2},2{{t}_{1}} \right)\And \left( {{t}_{2}}^{2},2{{t}_{2}} \right)$ respectively. Then using P and R points we will find one of the slopes of the two tangents and then using P and Q we will find the other slope of the tangent. Then we will get the values of ${{t}_{1}} \And {{t}_{2}}$. Hence, we will get the coordinates of Q and R.

Complete step by step solution:

The parabola given in the above problem is as follows:

${{y}^{2}}=4x$

From the point P (2, 3), two tangents are drawn to the above parabola so the two tangents will look like as follows:

Now, let us assume that the coordinates of point Q and point R are $\left( {{t}_{1}}^{2},2{{t}_{1}} \right)\And \left( {{t}_{2}}^{2},2{{t}_{2}} \right)$ respectively.

We know the equation of tangent to the parabola ${{y}^{2}}=4x$ is equal to $y=mx+\dfrac{a}{m}$. In the above problem, the value of $a=1$ so substituting this value of $''a''$ in the equation of the tangent and we get,

$y=mx+\dfrac{1}{m}$

The above tangent is passing through point P (2, 3) so substituting the values of x and y in the above equation we get,

$3=m2+\dfrac{1}{m}$

$\Rightarrow 3=2m+\dfrac{1}{m}$

Taking “m” as L.C.M in the above equation we get,

$3=\dfrac{2{{m}^{2}}+1}{m}$

On cross multiplying the above equation we get,

$3m=2{{m}^{2}}+1$

$\Rightarrow 2{{m}^{2}}-3m+1=0$

Writing 3 as (2 + 1) in the above equation we get,

$2{{m}^{2}}-\left( 2+1 \right)m+1=0$

$\Rightarrow 2{{m}^{2}}-2m-m+1=0$

Taking 2m as common from the first two terms and -1 from the last two terms we get,

$2m\left( m-1 \right)-1\left( m-1 \right)=0$

Taking $\left( m-1 \right)$ as common from the L.H.S of the above equation and we get,

$\left( m-1 \right)\left( 2m-1 \right)=0$

Equating each of the two brackets to 0 we get,

$m-1=0$

$\Rightarrow m=1;$

$2m-1=0$

$\Rightarrow m=\dfrac{1}{2}$

The “m” solved above is the slope of the tangent so first of all taking the slope 1 and using the points P and Q we get,

We know that we can write the slope using two points say $\left( {{x}_{1}},{{y}_{1}} \right)\And \left( {{x}_{2}},{{y}_{2}} \right)$ as follows: $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$

Now, using the above formula for points P (2, 3) and Q$\left( {{t}_{1}}^{2},2{{t}_{1}} \right)$ and we get,

$\dfrac{2{{t}_{1}}-3}{{{t}_{1}}^{2}-2}$

Now, equating the above slope to 1 we get,

$\dfrac{2{{t}_{1}}-3}{{{t}_{1}}^{2}-2}=1$

$\Rightarrow 2{{t}_{1}}-3={{t}_{1}}^{2}-2$

$\Rightarrow {{t}_{1}}^{2}-2{{t}_{1}}-2+3=0$

$\Rightarrow {{t}_{1}}^{2}-2{{t}_{1}}+1=0$

The above equation is in the form of the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ where $a={{t}_{1}}\And b=1$ so using this algebraic identity we get,

$\Rightarrow {{\left( {{t}_{1}}-1 \right)}^{2}}=0$

$\Rightarrow {{t}_{1}}=1$

Hence, we got the coordinates of point Q as (1, 2).

Now, we are going to find the slope using the points P (2, 3) and R $\left( {{t}_{2}}^{2},2{{t}_{2}} \right)$ so calculating the slope from these two points we get,

$\dfrac{2{{t}_{2}}-3}{{{t}_{2}}^{2}-2}$

Now, equating the above slope to $\dfrac{1}{2}$ we get,

$\dfrac{2{{t}_{2}}-3}{{{t}_{2}}^{2}-2}=\dfrac{1}{2}$

$\Rightarrow 2\left( 2{{t}_{2}}-3 \right)={{t}_{2}}^{2}-2$

$\Rightarrow 4{{t}_{2}}-6={{t}_{2}}^{2}-2$

$\Rightarrow {{t}_{2}}^{2}-2-4{{t}_{2}}+6=0$

$\Rightarrow {{t}_{2}}^{2}-4{{t}_{2}}+4=0$

The above equation is in the form of the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ where $a={{t}_{2}}\And b=2$ so using this algebraic identity we get,

$\begin{aligned}

$\Rightarrow {{\left( {{t}_{2}}-2 \right)}^{2}}=0$

$\Rightarrow {{t}_{2}}=2$

Hence, the coordinates of point R is equal to $\left( {{t}_{2}}^{2},2{{t}_{2}} \right)=\left( 4,4 \right)$.

So, the correct answer is “Option (b)”.

Note: To solve this problem, you have to have the knowledge of the equation of tangent to the parabola ${{y}^{2}}=4ax$. Also, you must know if two points are given then how we can find the slope from these two points. Failure of any of these two concepts will inhibit you to move forward in the above problem.