Question

The distance between the tangent lines to the hyperbola \[ x^2-2y^2=18 \] which are perpendicular to the line \(y=x\) is

• A. \(6\)
• B. \(3\sqrt{2}\)
• C. \(2\sqrt{3}\)
• D. \(0\)

Hint:

If a tangent line is perpendicular to a given line, first find its slope using \(m_1m_2=-1\). Then use the tangency condition by making the discriminant zero.

Answer 1

The Correct Option is B

Step 1: Find the slope of required tangent lines.
The given line is \[ y=x \] Its slope is \[ 1 \] Any line perpendicular to this line has slope \[ -1 \] So, the required tangent lines are of the form \[ y=-x+c \]

Step 2: Substitute the line in the hyperbola.
The hyperbola is \[ x^2-2y^2=18 \] Using \[ y=-x+c \] we get \[ x^2-2(-x+c)^2=18 \] Now, \[ (-x+c)^2=(x-c)^2=x^2-2cx+c^2 \] Therefore, \[ x^2-2(x^2-2cx+c^2)=18 \] \[ x^2-2x^2+4cx-2c^2=18 \] \[ -x^2+4cx-2c^2-18=0 \] Multiplying by \(-1\), \[ x^2-4cx+2c^2+18=0 \]

Step 3: Apply tangency condition.
For the line to be tangent to the hyperbola, the quadratic equation must have equal roots.
Hence, discriminant should be zero: \[ D=0 \] Here, \[ D=(-4c)^2-4(1)(2c^2+18) \] \[ D=16c^2-8c^2-72 \] \[ D=8c^2-72 \] Now, \[ 8c^2-72=0 \] \[ 8c^2=72 \] \[ c^2=9 \] \[ c=\pm 3 \] So, the tangent lines are \[ y=-x+3 \] and \[ y=-x-3 \] These can be written as \[ x+y-3=0 \] and \[ x+y+3=0 \]

Step 4: Find the distance between the parallel tangent lines.
The distance between two parallel lines \[ ax+by+c_1=0 \] and \[ ax+by+c_2=0 \] is \[ \frac{|c_1-c_2|}{\sqrt{a^2+b^2}} \] Here, the lines are \[ x+y-3=0 \] and \[ x+y+3=0 \] Thus, \[ d=\frac{| -3-3 |}{\sqrt{1^2+1^2}} \] \[ d=\frac{6}{\sqrt{2}} \] \[ d=3\sqrt{2} \]

Step 5: Final conclusion.
Therefore, the distance between the tangent lines is \[ \boxed{3\sqrt{2}} \]