Ellipse
1.0 Definition
2.0 Standard equation of Ellipse
3.0 Important terms
4.0 Difference between two forms of Ellipse
5.0 Focal Distance of a point
6.0 Parametric Co-ordinates
7.0 Equation of Tangent to Ellipse
7.1 Equation of tangent to a point/Point form
7.2 Parametric form
7.3 Equation of tangent in terms of slope/Slope form
8.0 Equation of Normal to Ellipse
9.0 Pair of tangents
10.0 Chord of contact
11.0 Chord bisected at a given point
12.0 Director circle
7.3 Equation of tangent in terms of slope/Slope form
7.2 Parametric form
7.3 Equation of tangent in terms of slope/Slope form
The equation of tangents of slope $m$ to ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ are $$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} $$
Coordinates of point of contact are $( \mp \frac{{{a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }}, \pm \frac{{{b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }})$.
Proof: Let $y = mx + c$ be the tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$. Put value of $y$ from tangent equation in ellipse equation as it touches the ellipse at a point and satisfies it. We get, $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{{(mx + c)}^2}}}{{{b^2}}} = 1$$
or, $${x^2}({a^2}{m^2} + {b^2}) + 2{a^2}mcx + {a^2}({c^2} - {b^2}) = 0\ ...(1)$$
Equation $(1)$ must have equal roots, therefore, ${B^2} - 4AC = 0$.
or, $$\begin{equation} \begin{aligned} 4{a^4}{m^2}{c^2} - 4({a^2}{m^2} + {b^2}){a^2}({c^2} - {b^2}) = 0 \\ {a^2}{m^2}{c^2} - ({a^2}{m^2} + {b^2})({c^2} - {b^2}) = 0 \\ {a^2}{m^2}{c^2} - {a^2}{m^2}{c^2} + {a^2}{b^2}{m^2} - {b^2}{c^2} + {b^4} = 0 \\ {a^2}{b^2}{m^2} - {b^2}{c^2} + {b^4} = 0 \\ {c^2} = {a^2}{m^2} + {b^2} \\ c = \pm \sqrt {{a^2}{m^2} + {b^2}} \\\end{aligned} \end{equation} $$
Substituting value of $c$ in $y = mx + c$, we get equation of tangent of the ellipse i.e., $$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} $$
By putting the value of $c$ in equation $(1)$, we get $$\begin{equation} \begin{aligned} {x^2}({a^2}{m^2} + {b^2}) \pm 2{a^2}m\sqrt {{a^2}{m^2} + {b^2}} x + {a^4}{m^2} = 0 \\ x = \mp \frac{{{a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }} \\\end{aligned} \end{equation} $$
Substituting value of $x$ in $y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} $, we obtained $$y = \pm \frac{{{b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }}$$
Question 9. Find the locus of point of intersection of perpendicular tangents to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$.
Solution: Let us assume the point of intersection be $P(h,k)$ and the equation of tangent to ellipse in terms of slope is $$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} $$
The point $P$ satisfies the equation of tangent i.e., $$k - mh = \pm \sqrt {{a^2}{m^2} + {b^2}} $$
Squaring both sides, we get $$\begin{equation} \begin{aligned} {k^2} + {m^2}{h^2} - 2mhk = {a^2}{m^2} + {b^2} \\ {m^2}({h^2} - {a^2}) - 2khm + {k^2} - {b^2} = 0 \\\end{aligned} \end{equation} $$
The slopes of tangent be ${m_1}$ and ${m_2}$ and $${m_1}{m_2} = \frac{{{k^2} - {b^2}}}{{{h^2} - a}}$$
As the two tangents are perpendicular, ${m_1}{m_2}=-1$
$$\begin{equation} \begin{aligned} - 1 = \frac{{{k^2} - {b^2}}}{{{h^2} - {a^2}}} \\ {a^2} - {h^2} = {k^2} - {b^2} \\ {h^2} + {k^2} = {a^2} + {b^2} \\\end{aligned} \end{equation} $$
Therefore, the locus of point of intersection of perpendicular tangents is $${x^2} + {y^2} = {a^2} + {b^2}$$
