Question

Consider the curve ${y^2} = 2x$ and point $A\left( {2,2} \right)$. If the normal at A intersects the curve again at point B and the tangent at A intersects the x-axis at C, then the area of $\Delta ABC$ is

Answer 1

Hint: In this question, we are given the equation of parabola ${y^2} = 2x$. First compare it with general equation of parabola and you’ll get the value of $a$. Now, find the equation of tangent at point A using tangent formula $y{y_1} = 2a\left( {x + {x_1}} \right)$. Now, it is given that tangent at A intersects the x-axis at C, put $y = 0$in the equation of tangent. You will get the coordinates of B. To find the coordinates of C, first find the equation of normal at point A using $\left( {y - {y_1}} \right) = \dfrac{{ - {y_1}}}{{2a}}\left( {x - {x_1}} \right)$ and then solve the required equation and given equation of curve. In last apply the formula of area of triangle $ = \dfrac{1}{2} \times base \times height$ and to calculate the sides use distance formula $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $.

Formula used:
General equation of parabola, ${y^2} = 4ax$
Equation of tangent to a parabola, $y{y_1} = 2a\left( {x + {x_1}} \right)$
Equation of normal to the parabola ${y^2} = 4ax$at point $\left( {{x_1},{y_1}} \right)$is –
$\left( {y - {y_1}} \right) = \dfrac{{ - {y_1}}}{{2a}}\left( {x - {x_1}} \right)$
Distance formula –
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ where $\left( {{x_1},{y_1}} \right),\left( {{y_1},{y_2}} \right)$ are the coordinates of first and second point of the line
Area of triangle$ = \dfrac{1}{2} \times base \times height$

Complete step by step solution:
Given equation of the curve is a parabola i.e., ${y^2} = 2x - - - - - \left( 1 \right)$
Compare equation (1) with general equation of parabola ${y^2} = 4ax$
It implies that $a = \dfrac{1}{2}$
Now using equation of tangent to a parabola, $y{y_1} = 2a\left( {x + {x_1}} \right)$
Therefore, the equation of tangent at point $A\left( {2,2} \right)$ will be $2y = x + 2 - - - - - \left( 2 \right)$
Also, the tangent at A intersects the x-axis at C (given)
Which implies that, $y = 0$
Put the value of $y = 0$in equation (2),
$ \Rightarrow x = - 2$
The co-ordinates of C are $\left( { - 2,0} \right)$
Now, using formula of equation of Normal i.e., $\left( {y - {y_1}} \right) = \dfrac{{ - {y_1}}}{{2a}}\left( {x - {x_1}} \right)$
Equation of normal at A will be $2x + y = 6 - - - - - \left( 3 \right)$
Now, it is given that the normal at A intersects the curve at point B
On solving equation (1) and (3),
$x = \dfrac{9}{2},y = - 3$
The co-ordinates of point B are $\left( {\dfrac{9}{2}, - 3} \right)$
Using distance formula,
$AC = \sqrt {{{\left( { - 2 - 2} \right)}^2} + {{\left( {0 - 2} \right)}^2}} $
$ = \sqrt {20} $
$AB = \sqrt {{{\left( {\dfrac{9}{2} - 2} \right)}^2} + {{\left( { - 3 - 2} \right)}^2}} $
$ = \sqrt {\dfrac{{25}}{4} + 25} $
Area of $\Delta ABC$$ = \dfrac{1}{2} \times base \times height$
$ = \dfrac{1}{2} \times AC \times AB$
$ = \dfrac{1}{2} \times \sqrt {20} \times \sqrt {\dfrac{{25}}{4} + 25} $
$ = \dfrac{{25}}{2}$
Hence, the area of the $\Delta ABC$ is $\dfrac{{25}}{2}$.

Note: A parabola is a U-shaped plane curve in which any point is an equal distance from both a fixed point (also known as the focus) and a fixed straight line (known as the directrix). A parabola is a right circular cone sectioned by a plane parallel to the cone's generator. Also, the normal to a parabola is perpendicular to the parabola's tangent. It will also pass through the point of intersection of the parabola's tangent.