Question

The equation of tangent with slope $m$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by $y=mx\mp \sqrt{a^2+b^2{m}^2}.$

• A. True
• B. False

Answer 1

The correct option is B

False

Any line with slope $m$ can be written $y=mx+c$. If we substitute $y=mx+c$ in the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ we will get a quadratic. Since $y=mx+c$ is a tangent, the quadratic should have only one solution, because tangent and the ellipse will intersect at only one point.

$\Rightarrow \frac{x^2}{a^2}+\frac{(mx+c{)}^2}{b^2}=1$ has one solution or one root.

We will simplify this quadratic and say discriminant = 0

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

$b^2{x}^2+a^2(m^2{x}^2+2mcx+c^2{)}^2=a^2{b}^2$

$(b^2+a^2{m}^2)x^2+2mc{a}^{2}x+a^2{c}^2-a^2{b}^2=0$

$△=0\Rightarrow (2mc{a}^2{)}^2-4(b^2+a^2{m}^2\left)a^2\right(c^2-b^2)=0$

$\Rightarrow 4m^{2}c2a^4-4a^2(b^2+a^2{m}^2)(c^2-b^2)=0$

$\Rightarrow m^2{c}^2{a}^2-b^2{c}^2-a^2{m}^2{c}^2+b^4+b^2{a}^2{m}^2)=0$

$\Rightarrow b^2{c}^2+b^4+b^2{a}^2{m}^2=0$

$\Rightarrow -c^2+b^2+a^2{m}^2=0$

$\Rightarrow c^2=b^2+a^2{m}^2$

or $c=\mp \sqrt{b^2+a^2{m}^2}$