Intersection of Line and a Parabola

Line and a Parabola

Lines are figures that are made up of infinite points extending indefinitely in both directions.

A line may meet the parabola in one point or two distinct points or it may not meet the parabola at all.

To get the point(s) of intersection, let us solve the equations of the parabola and the line simultaneously

Consider the standard equation of parabola $y^2=4 a x$ and the line having equation $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ Parabola is $\mathrm{y}^2=4 \mathrm{ax}$ and a line $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ then, $y^2=4 a\left(\frac{y-c}{m}\right)$

$
\Rightarrow m y^2-4 a y+4 a c=0
$
The above equation is quadratic in $y$

Depending on the discriminant of this equation, if we have 2 real roots, then 2 dis If we have 2 equal roots, then we have only one point where line touches the para If we do not have any real roots, then line does not intersect the parabola

Case 1: If the line meets the parabola in two distinct points $(R$ and $Q)$ the equation has two distinct real roots.

$
D \geqslant 0 \text { or } c<a / m
$
Case 2: If the line meets the parabola in one point (P), i.e., touches the parabola then the equation has two equal roots.

$
D=0 \text { or } c=a / m
$
Case 3: If the line doesn't meet the parabola then the equation has imaginary roots.

$
D<0 \text { or } c>a / m
$

Condition of tangency: The line $y=m x+c$ will be a tangent to the parabola $y^2=4 a x$, if $D=0 \Rightarrow c=$ $\mathrm{a} / \mathrm{m}$

Point of contact

Substitute the value of $c=a / m$ in the equation $m y^2-4 a y+4 a c=0$

$
\begin{array}{rlrl}
m y^2-4 a y+4 a\left(\frac{a}{m}\right) =0 \\
\Rightarrow m^2 y^2-4 a m y+4 a^2 =0 \\
\Rightarrow (m y-2 a)^2 =0 \\
\Rightarrow m y-2 a & =0 \text { or } y=\frac{2 a}{m}
\end{array}
$

Now, substitute the value of ' $y^{\prime}$ in $y=m x+\frac{a}{m}$ wre get, $\mathrm{x}=\frac{\mathrm{a}}{\mathrm{m}^2}$
Hence, point of contact is $\left(\frac{\mathrm{a}}{\mathrm{m}^2}, \frac{2 \mathrm{a}}{\mathrm{m}}\right)$

Recommended Video Based on Line and Parabola

Solved Examples Based on Line and Parabola

Example 1: The parabolas: $a x^2+2 b x+c y=0$ and $d x^2+2 e x+f y=0$ intersect on the line $y=1$ If $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f}$ are positive real numbers and $a, b, c$ are in G.P., then
[JEE MAINS 2023]
Solution: At $y=1$, Both curves intersect
$\left.\Rightarrow \begin{array}{l}a x^2+2 b x+c=0 \\ d x^2+2 \mathrm{ex}+\mathrm{f}=0\end{array}\right\}$ Common Root
Given $\mathrm{a}, \mathrm{b}$, and c are in G.P
$b^2=a c$
$\Rightarrow D=4 b^2-4 a c=0$ for the first equation
$\Rightarrow$ Both the Root are equal
$\because$ sum of the roots $=-2 \frac{b}{a}$

$\begin{aligned} & \alpha+\alpha=-2 \frac{b}{a} \\ & \alpha=-\frac{b}{a}\end{aligned}$
It satisfies the second equation also

$\begin{aligned} & d\left(-\frac{b}{a}\right)^2+2 e\left(-\frac{b}{a}\right)+f=0 \\ & d\left(\frac{b^2}{a^2}\right)-\frac{2 e b}{a}+f=0 \\ & d\left(\frac{a c}{a^2}\right)-2 e \frac{b}{a}+f=0 \\ & \frac{d}{a}-\frac{2 e b}{a c}+\frac{f}{c}=0 \\ & \frac{d}{a}-\frac{2 e b}{b^2}+\frac{f}{c}=0 \Rightarrow 2 \frac{e}{b}=\frac{d}{a}+\frac{f}{c} \Rightarrow \frac{d}{a}, \frac{e}{b}, \frac{f}{c} \text { are in } \mathrm{AP}\end{aligned}$

Hence, the answer is $\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}$ are in the AP

Example 2: If the line $\mathrm{y}=4+\mathrm{kx}, \mathrm{k}>0$, is the tangent to the parabola $\mathrm{y}=\mathrm{x}-\mathrm{x}^2$ at the point P and V is the vertex of the parabola, then the slope of the line through P and V is :
[JEE MAINS 2022]
Solution

$
\begin{aligned}
& y=k x+4 \\
& y=x-x^2
\end{aligned}
$
$
\begin{aligned}
& k x+4=x-x^2 \\
& x^2+(k-1) x+4=0 \\
& (k-1)^2-4 \cdot 4=0 \\
& k-1= \pm 4 \\
& \mathrm{f} k=5
\end{aligned}
$
Now put the value of $\mathrm{k}=5$

$
\begin{aligned}
& 5 x+4=x-x^2 \\
& x^2+4 x+4=0 \\
& (x+2)^2=0 \\
& x=-2 \\
& y=-6 \\
& \text { ff } k=-3
\end{aligned}
$
Now put the value of $\mathrm{k}=-3$ in eqn (1)

$
\begin{aligned}
& -3 x+4=x-x^2 \\
& x^2-4 x+4=0 \\
& x=2 \quad y=-2
\end{aligned}
$

Then the point of P is $(2,-2)$ and $(-2,-6)$ and vertex of the parabola

$
\begin{aligned}
& \mathrm{O}^{\prime}=\mathrm{y}-\frac{1}{4}=-\frac{1}{4}+\mathrm{x}-\mathrm{x}^2 \\
& \mathrm{y}-\frac{1}{4}=-\left(\mathrm{x}-\frac{1}{2}\right)^2
\end{aligned}
$
Point $P$ is $(2,2)$

Slope of

$
\mathrm{OP}=\frac{-2-\frac{1}{4}}{2-\frac{1}{2}}=\frac{-3}{2}
$
Point $P$ is $(-2,-6)$ the slope of

$
\mathrm{OP}=\frac{-6-\frac{1}{4}}{-2-\frac{1}{2}}=\frac{5}{2}
$
Hence, the answer is $\frac{5}{2}$

Example 3: If line $x+y=a$ and $x-y=b$ touch the curve $y=x^2-3 x+2$ at the pts where the curve intersects the $x$-axis then $a / b=$ ?
[JEE MAINS 2020]
Solution: Given the equation of the Curve $y=x^2-3 x+2$
The curve intersects at X -axis $(1,0)$ and $(2,0)$
Now the curve $\mathrm{x}+\mathrm{y}=\mathrm{a}$ and $\mathrm{x}-\mathrm{y}=\mathrm{b}$ touch the curve $y=x^2-3 x+2$
at $(1,0)$ and $(2,0)$
$a=1$ and $b=2$
$a / b=1 / 2$

Hence, the answer is the 1/2

Solution

$\begin{aligned} & P=\left(3 t^2, 6 t\right), N=\left(3 t^2, 0\right) \\ & M=\left(3 t^2, 3 t\right), Q=\left(\frac{3}{4} t^2, 3 t\right) \\ & \therefore t=\frac{1}{3} \\ & \therefore \mathrm{MQ}=\frac{9}{4} \mathrm{t}^2=\frac{1}{4} \text { and } \mathrm{PN}=6 \mathrm{t}=2\end{aligned}$

Hence, the answer is MQ = 1/4

Example 5: Events $A, B, C$ are mutually exclusive events such that $P(A)=\frac{3 x+1}{3}, \quad P(B)=\frac{1-x}{4} \quad$ and $P(C)=\frac{1-2 x}{2}$. The set of possible values of x is in the interval

Solution: Probability of occurrence of an event -Let $S$ be the sample space then the probability of occurrence of an event $E$ is denoted by $P(E)$ and it is defined as

$
\begin{aligned}
& P(E)=\frac{n(E)}{n(S)} \\
& P(E) \leq 1 \\
& P(E)=\lim _{n \rightarrow \infty}\left(\frac{r}{n}\right)
\end{aligned}
$

Where n repeated experiment and E occurs r times.

$
\begin{aligned}
& 0 \leq P(A) \leq 1 \\
& 0 \leq P(B) \leq 1 \\
& 0 \leq P(C) \leq 1 \\
& P(A)+P(B)+P(C) \leq 1\{\text { Conditions on probability value }
\end{aligned}
$

$
\text { Thus } 0 \leqslant \frac{3 x+1}{3} \leqslant 1
$
$
\Rightarrow-\frac{1}{3} \leqslant x \leqslant \frac{2}{3}
$

$1 a$

$
0 \leq \frac{1-x}{4} \leq 1
$
$
\Rightarrow-3 \leqslant x \leqslant 1
$
$
0 \leqslant \frac{1-2 x}{2} \leqslant 1
$

$
\Rightarrow \frac{-1}{2} \leqslant x \leqslant \frac{1}{2}
$
$
\begin{aligned}
& P(A)+P(B)+P(C) \leqslant 1 \\
& \Rightarrow \frac{3 x+1}{3}+\frac{1-x}{4}+\frac{1-2 x}{2} \leq 1 \\
& \Rightarrow 4(3 x+1)+3(1-x)+6(1-2 x) \leqslant 1 \\
& \Rightarrow-3 x+0 \leqslant-1
\end{aligned}
$

$\begin{aligned}
&\Rightarrow x \geq \frac{1}{3}\\
&\text { (4) }\\
&\text { From 1a,2a,3a and } 4 \text { we have }\\
&\frac{1}{3} \leqslant x \leqslant \frac{1}{2}
\end{aligned}$

Hence, the answer is $\left[\frac{1}{3}, \frac{1}{2}\right]$