Question

The straight line $y = m(x - a)$ will meet the parabola ${y^2} = 4ax$ in two distinct points, if
a.$m \in R$
b.$m \in [ - 1,1]$
c.$m \in [ - \infty , - 1] \cup [1,\infty ]$
d.$m \in R - \{ 0\} $

Answer 1

Hint: For finding the point of intersection of two curves, we have to use a substitution method using the 2 equations. Here we have to find the range of values of m.

Complete step-by-step answer:
The given equation of straight line and parabola respectively is
$y = m(x - a)$ …………….. 1
${y^2} = 4ax$ …………………. 2
For points of intersection, we have to substitute the value of $y$ in terms of $x$ in the 2nd equation.
$ \Rightarrow {[m(x - a)]^2} = 4ax$
$ \Rightarrow {m^2}{(x - a)^2} - 4ax = 0$
$ \Rightarrow {m^2}({x^2} + a - 2ax) - 4ax = 0$
$ \Rightarrow {m^2}{x^2} + {m^2}{a^2} - 2{m^2}ax - 4ax = 0$
$ \Rightarrow {m^2}{x^2} + {m^2}{a^2} - 2ax({m^2} + 2) = 0$ ………………… 3
Equation 3 is a quadratic equation, thus for a quadratic equation to have two real and distinct roots
${b^2} - 4ac > 0$
$ \Rightarrow {[2a({m^2} + 2)]^2} - 4({m^2})({m^2}{a^2}) > 0$
$ \Rightarrow 4{a^2}({m^4} + 4{m^2} + 4) - 4{m^4}{a^2} > 0$
$ \Rightarrow 4{a^2}{m^4} + 16{a^2}{m^2} + 16{a^2} - 4{m^4}a > 0$
$ \Rightarrow 16{a^2}{m^2} + 16{a^2} > 0$
$ \Rightarrow 16{a^2}({m^2} + 1) > 0$ ……………… 4
Hence the given straight line will intersect the parabola at 2 distinct points if the above expression 4 is true.
Now, for any value of $m \in R$ , expression 4 will always be true thus the line will cut parabola for all values of $m \in R$.
But at m=0, y=0 $ \Rightarrow $no parabola in such case, so it will not be counted.
And hence the straight line will intersect the parabola for all values of $m \in R - \{ 0\} $.
Option D is correct answer.

Note: For a quadratic expression,remember these formula
1.For real and distinct roots,
${b^2} - 4ac > 0$
2.For real and equal roots,
${b^2} - 4ac = 0$
3.For imaginary roots,
${b^2} - 4ac < 0$