Question

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the straight line $y=mx+c$ intersect in real points only, if?

• A. $a^2{m}^2<c^2-b^2$
• B. $a^2{m}^2>c^2+b^2$
• C. $a^2{m}^2\ge c^2-b^2$
• D. $c\ge b$

Answer 1

The correct option is A $a^2{m}^2<c^2-b^2$

$\frac{x^2}{a^2}+\frac{{(mx+c)}^2}{b^2}=1$

$\Rightarrow \frac{x^2}{a^2}+\frac{m^2{x}^2}{b^2}+\frac{c^2}{b^2}+\frac{2cmn}{b^2}=1$

$\Rightarrow b^2{x}^2+a^2(m^2{x}^2+2cmn+c^2)=a^2{b}^2$

$\Rightarrow x^2(b^2+a^2{m}^2)+x\left({2cma}^2\right)+a^2{c}^2-a^2{b}^2=0$

for $x$ to be real,

${\left({2cma}^2\right)}^2\ge 4(b^2+a^2{m}^2)(a^2{c}^2-a^2{b}^2)$

$\Rightarrow {4c}^2{m}^2{a}^4\ge {4a}^2{b}^2{c}^2-a^2{b}^4+a^4{m}^2{c}^2-a^4{b}^2{m}^2$

$\Rightarrow a^2{c}^2{m}^2\ge b^2{c}^2-b^4+a^2{m}^2{c}^2-a^2{b}^2{m}^2$

$\Rightarrow 0\ge c^2-b^2-a^2{m}^2\Rightarrow a^2{m}^2\ge c^2-b^2$ [C]