Author Topic: conics (Read 1408 times) Tweet Share

massive · « **on:** July 12, 2016, 04:32:31 pm »

hey guys howdu these questions???

Show that the condition for the line y=mx + c to be tangent to the ellipse x^2/a^2 +y^2/b^2 =1 is c^2=a^2m^2 + b^2

show that the pair of tangents drawn from the point (3,4) to the ellipse x^2/16 + y^2/9 = 1 are at right angles to one another

RuiAce · « **Reply #1 on:** July 12, 2016, 08:10:52 pm »

Quote from: massive on July 12, 2016, 04:32:31 pm

hey guys howdu these questions???

Show that the condition for the line y=mx + c to be tangent to the ellipse x^2/a^2 +y^2/b^2 =1 is c^2=a^2m^2 + b^2

show that the pair of tangents drawn from the point (3,4) to the ellipse x^2/16 + y^2/9 = 1 are at right angles to one another

$\text{We request that the line must be tangential to the ellipse. This directly implies a point of intersection exists.}\\ \text{Suppose we want to find the point of intersection. We are then solving (simultaneously):}\\ \begin{align}y&=mx+c\\ \frac{x^2}{a^2}+\frac{y^2}{b^2}&=1\end{align}$
$\text{Substitute the first equation into the second:}\\ \begin{align*}b^2x^2+a^2y^2&=a^2b^2\\ \therefore b^2x^2+a^2(m^2x^2+2mcx+c^2)&=a^2b^2\\ (b^2+a^2m^2)x^2+(2a^2mc)x+(a^2c^2-a^2b^2)&=0\end{align*}$
$\textbf{If a line is tangential to a conic section, then there exists EXACTLY one UNIQUE point of intersection.}\\ \textbf{This is because a tangent to a conic never cuts the conic ever again.}\\ \text{Thus the above equation has only one solution. As it is quadratic in }x\text{, the polynomial discriminant }\Delta = 0$
$\begin{align*}(2a^2mc)^2 - 4(b^2+a^2m^2)(a^2c^2-a^2b^2)&=0 \\ (4a^4m^2c^2)-4(a^2b^2c^2-a^2b^4+a^4c^2m^2-a^4b^2m^2) &=0 \\ 4(a^2b^4-a^2b^2c^2+a^4b^2m^2)&=0\\ c^2&=a^2m^2+b^2 \end{align*}\\ \text{noting }a\neq 0, b\neq 0$

massive · « **Reply #2 on:** July 12, 2016, 08:24:17 pm »

Wow!! thanks you beast, also howdu do the second part? :S

RuiAce · « **Reply #3 on:** July 12, 2016, 08:48:40 pm »

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Author Topic: conics (Read 1408 times) Tweet Share

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