Question

If y=x is a tangent to the parabola $y=a{{x}^{2}}+c$ and $c=2$, then find the point of contact.
(a) (3, 3)
(b) (2, 2)
(c) (6, 6)
(d) (4, 4)

Answer 1

Hint: Here, first we need to find the slope of the line tangent to the parabola and equate it with the slope of the parabola, then consider the point of contact be P (h, k) and substitute the point in the equation of tangent $\left( x,\,y \right)$. Substitute the (h, k) in the equation of the parabola and find the value of k. With the help of the equation of tangent equate h and k and find the value of $a$. Now, finally, substitute the value of $a$ in h and find the point of contact.

Complete step-by-step solution
We have been given the equation of tangent $y=x$ which touches the parabola at $y=a{{x}^{2}}+c$ and we have $c=2$.
First let us differentiate the equation of tangent with respect to $x$.
$\dfrac{dy}{dx}=1$
Now, substitute the
$y=a{{x}^{2}}+2$
Let us also differentiate the equation of parabola with respect to $x$.
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{d}{dx}\left( a{{x}^{2}}+2 \right) \\
 & =2ax
\end{align}$
Let the point of contact be P (h, k) when the tangent touches the parabola, therefore we can say that, $\left( x,\,y \right)=$P (h, k)
Now, we know that the slope $m=\dfrac{dy}{dx}$.
Slope of tangent $=1$
Slope of the parabola $=2a\text{h}$
Here, since the line equation is tangent to the parabola, we can say that the slope of the tangent is equal to slope of the parabola, we get
\[\begin{align}
  & 1=2a\text{h} \\
 & \text{h =}\dfrac{1}{2a} \\
\end{align}\]
Since, $\left( x,\,y \right)=$P (h, k), we can rewrite the $y=a{{x}^{2}}+2$ as $\text{k = }a{{h}^{2}}+2$.
Now, after substituting the value of h in the above obtained expression, we get
$\begin{align}
  & \text{k =}\,a{{\left( \dfrac{1}{2a} \right)}^{2}}+2 \\
 & =a\left( \dfrac{1}{4{{a}^{2}}} \right)+2 \\
 & =\dfrac{1}{4a}+2 \\
 & =\dfrac{1+2\left( 4a \right)}{4a} \\
 & =\dfrac{1+8a}{4a}
\end{align}$
Therefore, we have $\left( \text{h, k} \right)=\,\left( \dfrac{1}{2a},\,\dfrac{1+8a}{4a} \right)$.
From the equation of tangent $y=x$ we can say that h = k, so we get
$\begin{align}
  & \dfrac{1}{2a}=\dfrac{1+8a}{4a} \\
 & 1=\dfrac{1+8a}{2} \\
 & 2=1+8a \\
 & 2-1=8a \\
 & 1=8a \\
 & a=\dfrac{1}{8}
\end{align}$
Since, h = k, we can say that (h, k) = (h, h)
Let us find the original equation of parabola by substituting the value of $a$ and the value of $c$ in the equation of parabola, we get
$y=\dfrac{1}{8}{{x}^{2}}+2$
Let us draw the figure of parabola with the above equation along with the tangent that touches it at a point P.

Now, let us find the value of h by substituting the value of $a$ in $\text{h}=\dfrac{1}{2a}$
$\begin{align}
  & \text{h =}\dfrac{1}{2\left( \dfrac{1}{8} \right)} \\
 & =\dfrac{1}{\dfrac{1}{4}} \\
 & =4
\end{align}$
Therefore, (h, k) = (h, h) = (4, 4).
Hence, the point of contact of the tangent to the parabola is P (4, 4).

Note: A parabola is a locus of the point which moves in a plane such that its distance from a fixed point (focus) is always equal to its distance from a fixed straight line (directrix). Here, in this question, since the point of contact lies on the tangent and on the parabola, we can substitute the point of contact (h, k) in both the equations of parabola and tangent. The diagram can be drawn to check whether the answer we got is correct.