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HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 2 => Topic started by: massive on July 12, 2016, 04:32:31 pm

Title: conics
Post by: 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
Title: Re: conics
Post by: RuiAce 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
$\text{We request that the line must be tangential to the ellipse. This directly implies a point of intersection exists.}\\ \text{Suppose we want to find the point of intersection. We are then solving (simultaneously):}\\ \begin{align}y&=mx+c\\ \frac{x^2}{a^2}+\frac{y^2}{b^2}&=1\end{align}$
$\text{Substitute the first equation into the second:}\\ \begin{align*}b^2x^2+a^2y^2&=a^2b^2\\ \therefore b^2x^2+a^2(m^2x^2+2mcx+c^2)&=a^2b^2\\ (b^2+a^2m^2)x^2+(2a^2mc)x+(a^2c^2-a^2b^2)&=0\end{align*}$
$\textbf{If a line is tangential to a conic section, then there exists EXACTLY one UNIQUE point of intersection.}\\ \textbf{This is because a tangent to a conic never cuts the conic ever again.}\\ \text{Thus the above equation has only one solution. As it is quadratic in }x\text{, the polynomial discriminant }\Delta = 0$
$\begin{align*}(2a^2mc)^2 - 4(b^2+a^2m^2)(a^2c^2-a^2b^2)&=0 \\ (4a^4m^2c^2)-4(a^2b^2c^2-a^2b^4+a^4c^2m^2-a^4b^2m^2) &=0 \\ 4(a^2b^4-a^2b^2c^2+a^4b^2m^2)&=0\\ c^2&=a^2m^2+b^2 \end{align*}\\ \text{noting }a\neq 0, b\neq 0$
Title: Re: conics
Post by: massive on July 12, 2016, 08:24:17 pm: Wow!! thanks you beast, also howdu do the second part? :S
Title: Re: conics
Post by: RuiAce on July 12, 2016, 08:48:40 pm: $\text{Applying the condition we've proven to this new question where }a=4,b=3:\\ y=mx+c\text{ is tangential to the ellipse }\frac{x^2}{16}+\frac{y^2}{9}=1\text{ if and only if}\\ c^2=16m^2+9 \implies c=\pm \sqrt{16m^2+9}$
$\text{IF this condition is satisfied, then the two tangents are}\\ y=mx\pm\sqrt{16m^2+9}$
$\text{We know that }(3,4)\text{ passes through both tangents.}\\ \begin{align*} \therefore 4&=3m\pm \sqrt{16m^2+9} \\ (4-3m)^2&=(\pm \sqrt{16m^2+9})^2\\ 16-24m+9m^2&=16m^2+9\\ 7m^2 + 24m - 7 &= 0\end{align*}$
$\text{If you think about it, the solutions }m=m_1, m=m_2\text{ are going to be the gradients of}\textbf{ both }\text{tangents at }(3,4)!$
$\text{By the product of roots: }m_1m_2=\frac{7}{-7}=-1$
$\text{This perfectly satisfies the condition for the tangents to be perpendicular as required.}$