Question

If the focus is \[\left( 1,2 \right)\]and the equation of directrix is \[7x-y-30=0\]and the latus rectum is 10 units then the conic is
(a) Parabola
(b) Ellipse
(c) Hyperbola
(d) Rectangular hyperbola

Answer 1

Hint: We solve this problem by using the relation between focus, directrix, and latus rectum.
We have the formula that if \[d\] is the distance between focus and directrix, \[e\]is the eccentricity of the conic and \[l\] is the latus rectum then
\[\Rightarrow l=2ed\]
We have the formula of distance from \[\left( {{x}_{1}},{{y}_{1}} \right)\] to line \[ax+by+c=0\] is given as
\[d=\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|\]
By using the above conditions we find the eccentricity of the conic.
(1) If \[e=1\] then conic is parabola
(2) If \[e<1\] then conic is ellipse
(3) If \[e>1\] then conic is hyperbola
(4) If \[e=\sqrt{2}\] then conic is rectangular hyperbola.

Complete step by step answer:
We are given that the focus of conic is \[\left( 1,2 \right)\]
We are also given that the equation of directrix is given as \[7x-y-30=0\]
Let us assume that the distance between focus and directrix as \[d\]
We know that the formula of distance from \[\left( {{x}_{1}},{{y}_{1}} \right)\] to line \[ax+by+c=0\] is given as
\[d=\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|\]
By using the above formula we get the distance between focus and directrix as
\[\begin{align}
  & \Rightarrow d=\left| \dfrac{7-2-30}{\sqrt{{{7}^{2}}+{{1}^{2}}}} \right| \\
 & \Rightarrow d=\dfrac{25}{\sqrt{50}} \\
\end{align}\]
Here, we can see that the value of \[\sqrt{50}\] can be written as \[5\sqrt{2}\] then we get
\[\begin{align}
  & \Rightarrow d=\dfrac{25}{5\sqrt{2}} \\
 & \Rightarrow d=\dfrac{5}{\sqrt{2}} \\
\end{align}\]
We are given that the length of latus rectum as 10 units
We know that the formula that if \[d\] is the distance between focus and directrix, \[e\]is the eccentricity of the conic and \[l\] is the latus rectum then
\[\Rightarrow l=2ed\]
By using the above formula to given values we get
\[\begin{align}
  & \Rightarrow 10=2e\left( \dfrac{5}{\sqrt{2}} \right) \\
 & \Rightarrow e=\sqrt{2} \\
\end{align}\]
We know that the condition that if \[e=\sqrt{2}\] then conic is a rectangular hyperbola.
Therefore we can conclude that the given conic is rectangular hyperbola as follows

Note:
Students may make mistakes in taking the relation between the latus rectum and the distance between focus and directrix.
We have the formula that if \[d\] is the distance between focus and directrix, \[e\]is the eccentricity of the conic, and \[l\] is the latus rectum then
\[\Rightarrow l=2ed\]
Bit students may assume that the equation without the eccentricity as
\[\Rightarrow l=2d\]
This is correct when we know the conic is a parabola.
But here, we don’t know about the conic so that we cannot neglect the eccentricity.