The equation of tangent with slope m of the circle $x^{2} + y^{2} = a^{2}$ are given by $y = m x \pm m \sqrt{1 + a^{2}}$

True

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False

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Solution

The correct option is B
False

We want to find the equation of the tangent whose slope is m. the equation given to us is of the form y

=mx+c. we have to check if c= $\pm m \sqrt{1 + a^{2}}$

Let the equation of the circle be y=mx+c

We know the perpendicular from center to the tangent is equal to radius.

(we will be using this condition in solving problem related to tangents)

$\Rightarrow$ Distance of (0,0) from mx-y+c=0,=4

$\Rightarrow ∣ ∣ ∣ \frac{c}{\sqrt{1 + m^{2}}} ∣ ∣ ∣ = a$

$\Rightarrow \frac{c}{\sqrt{1 + m^{2}}} = \pm a$

$\Rightarrow c = \pm a \sqrt{1 + m^{2}}$

$\Rightarrow$ The given statement is wrong

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

The equation of the normal of slope m to the hyperbola
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
is given by $y = m x \pm \frac{m (a^{2} + b^{2})}{\sqrt{a^{2} - b^{2} m^{2}}}$

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

Q. What is the equation of the normal with slope m to the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 ?

A

line touches the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

and the circle

x^{2} + y^{2} = r^{2}

Then the slope

m

of the common tangent is given by

m^{2} =

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The equation of tangent with slope m of the circle x2+y2=a2 are given by y=mx±m√1+a2

The equation of tangent with slope m of the circle $x^{2} + y^{2} = a^{2}$ are given by $y = m x \pm m \sqrt{1 + a^{2}}$