# Find the condition under which the line xcosalpha+ysinalpha=p will be a tangent to the conic 3x^2+4y^2=5?

## Find the condition under which the line $x \cos \alpha + y \sin \alpha = p$ will be a tangent to the conic $3 {x}^{2} + 4 {y}^{2} = 5$

Mar 15, 2017

See below.

#### Explanation:

Given $f \left(x , y\right) = 3 {x}^{2} + 4 {y}^{2} - 5$ the normal to the curve $f \left(x . y\right) = 0$ at point ${p}_{0} = \left({x}_{0} , {y}_{0}\right)$ is

${\vec{n}}_{0} = \left({f}_{x} , {f}_{y}\right) \left({x}_{0} , {y}_{0}\right) = \left(6 {x}_{0} , 8 {y}_{0}\right) = 2 \left(3 {x}_{0} , 4 {y}_{0}\right)$

Calling $p = \left(x , y\right)$ a tangent to $f \left(x , y\right) = 0$ at point ${x}_{0} , {y}_{0}$ is given by

$l \to \left\langlep - {p}_{0} , {\vec{n}}_{0}\right\rangle = 0$

where $\left\langle\cdot , \cdot\right\rangle$ denotes the scalar product of two vectors, or

$l \to 6 x {x}_{0} + 8 y {y}_{0} - 2 \left(3 {x}_{0}^{2} + 4 {y}_{0}^{2}\right) = 0$ or

$l \to 6 x {x}_{0} + 8 y {y}_{0} - 2 \cdot 5 = 0$ or

$l \to 3 x {x}_{0} + 4 y {y}_{0} - 5 = 0$ and also

$l \to x \left(\frac{3 {x}_{0}}{\sqrt{{\left(3 {x}_{0}\right)}^{2} + {\left(4 {y}_{0}\right)}^{2}}}\right) + y \left(\frac{4 {y}_{0}}{\sqrt{{\left(3 {x}_{0}\right)}^{2} + {\left(4 {y}_{0}\right)}^{2}}}\right) = \frac{5}{\sqrt{{\left(3 {x}_{0}\right)}^{2} + {\left(4 {y}_{0}\right)}^{2}}}$

Now, calling

$\cos \alpha = \frac{3 {x}_{0}}{\sqrt{{\left(3 {x}_{0}\right)}^{2} + {\left(4 {y}_{0}\right)}^{2}}}$
$\sin \alpha = \frac{4 {y}_{0}}{\sqrt{{\left(3 {x}_{0}\right)}^{2} + {\left(4 {y}_{0}\right)}^{2}}}$
$p = \frac{5}{\sqrt{{\left(3 {x}_{0}\right)}^{2} + {\left(4 {y}_{0}\right)}^{2}}}$

or

$\cos \alpha = \frac{3 {x}_{0}}{\sqrt{20 - 3 {x}_{0}^{2}}}$
sin alpha = 2sqrt((3x_0^2-5)/(3x_0^2-20)
$p = \frac{5}{\sqrt{20 - 3 {x}_{0}^{2}}}$

we have the sough tangent line