JEE Advanced · Mathematics · 9. Straight Lines
The locus of the orthocentre of the triangle formed by the lines \((1+p) x-p y+p(1+p)=0\), \((1+q) x-q y+q(1+q)=0\) and \(y=0\), where \(p \neq q\), is
- A
a hyperbola
- B
a parabola
- C
an ellipse
- D
a straight line
Answer & Solution
Correct Answer
(D)
a straight line
Step-by-step Solution
Detailed explanation
Given, lines are
\[
(1+p) x-p y+p(1+p)=0
\]
and \((1+q) x-q y+q(1+q)=0\)
On solving Eqs. (i) and (ii), we get
\[
C\{p q,(1+p)(1+q)\}
\]
\(\therefore\) Equation of altitude \(C M\) passing through \(C\) and perpendicular to \(A B\) is
\[
x=p q
\]
\(\because\) Slope of line (ii) is \(\left(\frac{1+q}{q}\right)\).
\(\therefore\) Slope of altitude \(B N\) (as shown in figure) is \(\frac{-q}{1+q}\).

\(\therefore\) Equation of \(B N\) is \(y-0=\frac{-q}{1+q}(x+p)\)
\[
\Rightarrow \quad y=\frac{-q}{(1+q)}(x+p)
\]
Let orthocentre of triangle be \(H(h, k)\), which is the point of intersection of Eqs. (iii) and (iv)
On solving Eqs. (iii) and (iv), we get \(x=p q\) and \(y=-p q \Rightarrow h=p q\) and \(k=-p q \Rightarrow h+k=0\)
\(\therefore\) Locus of \(H(h, k)\) is \(x+y=0\).
\[
(1+p) x-p y+p(1+p)=0
\]
and \((1+q) x-q y+q(1+q)=0\)
On solving Eqs. (i) and (ii), we get
\[
C\{p q,(1+p)(1+q)\}
\]
\(\therefore\) Equation of altitude \(C M\) passing through \(C\) and perpendicular to \(A B\) is
\[
x=p q
\]
\(\because\) Slope of line (ii) is \(\left(\frac{1+q}{q}\right)\).
\(\therefore\) Slope of altitude \(B N\) (as shown in figure) is \(\frac{-q}{1+q}\).

\(\therefore\) Equation of \(B N\) is \(y-0=\frac{-q}{1+q}(x+p)\)
\[
\Rightarrow \quad y=\frac{-q}{(1+q)}(x+p)
\]
Let orthocentre of triangle be \(H(h, k)\), which is the point of intersection of Eqs. (iii) and (iv)
On solving Eqs. (iii) and (iv), we get \(x=p q\) and \(y=-p q \Rightarrow h=p q\) and \(k=-p q \Rightarrow h+k=0\)
\(\therefore\) Locus of \(H(h, k)\) is \(x+y=0\).
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