Question:
Let origin be the centre, \((\pm 3,0)\) be the foci and \(\dfrac{3}{2}\) be the eccentricity of a hyperbola. Then the line \(2x-y-1=0\)
Let origin be the centre, \((\pm 3,0)\) be the foci and \(\dfrac{3}{2}\) be the eccentricity of a hyperbola. Then the line \(2x-y-1=0\)
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To check whether a line intersects a conic, substitute the line equation into the conic and examine the discriminant of the resulting quadratic equation.
Updated On: Jun 15, 2026
- intersects the hyperbola at two points
- does not intersect the hyperbola
- touches the hyperbola
- passes through the vertex of the hyperbola
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The Correct Option is B
Solution and Explanation
Step 1: Find the equation of the hyperbola.
The centre of the hyperbola is the origin and its foci are \[ (\pm 3,0) \] So, the hyperbola has transverse axis along the \(x\)-axis.
Hence, its equation is of the form \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] Here, \[ c=3 \] The eccentricity is given as \[ e=\frac{3}{2} \] Since \[ e=\frac{c}{a}, \] we get \[ \frac{3}{2}=\frac{3}{a} \] Therefore, \[ a=2 \] So, \[ a^2=4 \]
Step 2: Find \(b^2\).
For a hyperbola, \[ c^2=a^2+b^2 \] Substituting the values, \[ 3^2=2^2+b^2 \] \[ 9=4+b^2 \] \[ b^2=5 \] Thus, the equation of the hyperbola is \[ \frac{x^2}{4}-\frac{y^2}{5}=1 \]
Step 3: Substitute the given line in the hyperbola.
The given line is \[ 2x-y-1=0 \] So, \[ y=2x-1 \] Substitute this in \[ \frac{x^2}{4}-\frac{y^2}{5}=1 \] \[ \frac{x^2}{4}-\frac{(2x-1)^2}{5}=1 \] Multiplying by \(20\), \[ 5x^2-4(2x-1)^2=20 \] \[ 5x^2-4(4x^2-4x+1)=20 \] \[ 5x^2-16x^2+16x-4=20 \] \[ -11x^2+16x-24=0 \] Multiplying by \(-1\), \[ 11x^2-16x+24=0 \]
Step 4: Check the discriminant.
For \[ 11x^2-16x+24=0, \] the discriminant is \[ D=b^2-4ac \] \[ D=(-16)^2-4(11)(24) \] \[ D=256-1056 \] \[ D=-800 \] Since \[ D\lt 0, \] the line does not meet the hyperbola at any real point.
Step 5: Final conclusion.
Therefore, the line \[ 2x-y-1=0 \] does not intersect the hyperbola.
\[ \boxed{\text{does not intersect the hyperbola}} \]
The centre of the hyperbola is the origin and its foci are \[ (\pm 3,0) \] So, the hyperbola has transverse axis along the \(x\)-axis.
Hence, its equation is of the form \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] Here, \[ c=3 \] The eccentricity is given as \[ e=\frac{3}{2} \] Since \[ e=\frac{c}{a}, \] we get \[ \frac{3}{2}=\frac{3}{a} \] Therefore, \[ a=2 \] So, \[ a^2=4 \]
Step 2: Find \(b^2\).
For a hyperbola, \[ c^2=a^2+b^2 \] Substituting the values, \[ 3^2=2^2+b^2 \] \[ 9=4+b^2 \] \[ b^2=5 \] Thus, the equation of the hyperbola is \[ \frac{x^2}{4}-\frac{y^2}{5}=1 \]
Step 3: Substitute the given line in the hyperbola.
The given line is \[ 2x-y-1=0 \] So, \[ y=2x-1 \] Substitute this in \[ \frac{x^2}{4}-\frac{y^2}{5}=1 \] \[ \frac{x^2}{4}-\frac{(2x-1)^2}{5}=1 \] Multiplying by \(20\), \[ 5x^2-4(2x-1)^2=20 \] \[ 5x^2-4(4x^2-4x+1)=20 \] \[ 5x^2-16x^2+16x-4=20 \] \[ -11x^2+16x-24=0 \] Multiplying by \(-1\), \[ 11x^2-16x+24=0 \]
Step 4: Check the discriminant.
For \[ 11x^2-16x+24=0, \] the discriminant is \[ D=b^2-4ac \] \[ D=(-16)^2-4(11)(24) \] \[ D=256-1056 \] \[ D=-800 \] Since \[ D\lt 0, \] the line does not meet the hyperbola at any real point.
Step 5: Final conclusion.
Therefore, the line \[ 2x-y-1=0 \] does not intersect the hyperbola.
\[ \boxed{\text{does not intersect the hyperbola}} \]
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