Question:
The number of those tangents to the curve
\[
y^2-2x^3-4y+8=0
\]
which pass through the point \((1,2)\) is
The number of those tangents to the curve
\[
y^2-2x^3-4y+8=0
\]
which pass through the point \((1,2)\) is
Show Hint
For finding tangents from an external point to an implicit curve, take a general point \((a,b)\) on the curve, find the slope using implicit differentiation, and equate it with the slope of the line joining \((a,b)\) to the given external point.
Updated On: Jun 22, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Rewrite the curve.
Given curve is \[ y^2-2x^3-4y+8=0 \] Rearranging the terms involving \(y\), \[ y^2-4y+8=2x^3 \] Completing the square, \[ (y-2)^2+4=2x^3 \] So, \[ (y-2)^2=2x^3-4 \]
Step 2: Check whether the given point lies on the curve.
Substitute \((1,2)\) in the curve: \[ 2^2-2(1)^3-4(2)+8 \] \[ =4-2-8+8 \] \[ =2\neq 0 \] Hence, \((1,2)\) does not lie on the curve.
Step 3: Find the tangent condition.
Let the tangent point on the curve be \((a,b)\). Since \((a,b)\) lies on the curve, \[ b^2-2a^3-4b+8=0 \] Differentiating implicitly, \[ 2y\frac{dy}{dx}-6x^2-4\frac{dy}{dx}=0 \] \[ (2y-4)\frac{dy}{dx}=6x^2 \] \[ \frac{dy}{dx}=\frac{3x^2}{y-2} \] Therefore, slope of tangent at \((a,b)\) is \[ m=\frac{3a^2}{b-2} \]
Step 4: Use condition that tangent passes through \((1,2)\).
The slope of line joining \((a,b)\) and \((1,2)\) is \[ \frac{b-2}{a-1} \] Since this line is tangent, \[ \frac{b-2}{a-1}=\frac{3a^2}{b-2} \] Thus, \[ (b-2)^2=3a^2(a-1) \] From the curve, \[ (b-2)^2=2a^3-4 \] Equating both expressions, \[ 2a^3-4=3a^2(a-1) \] \[ 2a^3-4=3a^3-3a^2 \] \[ a^3-3a^2+4=0 \] \[ (a-2)^2(a+1)=0 \] So, \[ a=2 \quad \text{or} \quad a=-1 \]
Step 5: Check valid tangent points.
For \(a=2\), \[ (b-2)^2=2(2)^3-4=16-4=12 \] So real values of \(b\) exist.
For \(a=-1\), \[ (b-2)^2=2(-1)^3-4=-2-4=-6 \] This is not possible for real \(b\).
Thus, only \(a=2\) gives real tangent points. Although two points are obtained for \(b\), both give the same tangent line passing through \((1,2)\).
Therefore, the number of tangents is \[ 1 \]
Step 6: Final conclusion.
Hence, the required number of tangents is \[ \boxed{1} \]
Given curve is \[ y^2-2x^3-4y+8=0 \] Rearranging the terms involving \(y\), \[ y^2-4y+8=2x^3 \] Completing the square, \[ (y-2)^2+4=2x^3 \] So, \[ (y-2)^2=2x^3-4 \]
Step 2: Check whether the given point lies on the curve.
Substitute \((1,2)\) in the curve: \[ 2^2-2(1)^3-4(2)+8 \] \[ =4-2-8+8 \] \[ =2\neq 0 \] Hence, \((1,2)\) does not lie on the curve.
Step 3: Find the tangent condition.
Let the tangent point on the curve be \((a,b)\). Since \((a,b)\) lies on the curve, \[ b^2-2a^3-4b+8=0 \] Differentiating implicitly, \[ 2y\frac{dy}{dx}-6x^2-4\frac{dy}{dx}=0 \] \[ (2y-4)\frac{dy}{dx}=6x^2 \] \[ \frac{dy}{dx}=\frac{3x^2}{y-2} \] Therefore, slope of tangent at \((a,b)\) is \[ m=\frac{3a^2}{b-2} \]
Step 4: Use condition that tangent passes through \((1,2)\).
The slope of line joining \((a,b)\) and \((1,2)\) is \[ \frac{b-2}{a-1} \] Since this line is tangent, \[ \frac{b-2}{a-1}=\frac{3a^2}{b-2} \] Thus, \[ (b-2)^2=3a^2(a-1) \] From the curve, \[ (b-2)^2=2a^3-4 \] Equating both expressions, \[ 2a^3-4=3a^2(a-1) \] \[ 2a^3-4=3a^3-3a^2 \] \[ a^3-3a^2+4=0 \] \[ (a-2)^2(a+1)=0 \] So, \[ a=2 \quad \text{or} \quad a=-1 \]
Step 5: Check valid tangent points.
For \(a=2\), \[ (b-2)^2=2(2)^3-4=16-4=12 \] So real values of \(b\) exist.
For \(a=-1\), \[ (b-2)^2=2(-1)^3-4=-2-4=-6 \] This is not possible for real \(b\).
Thus, only \(a=2\) gives real tangent points. Although two points are obtained for \(b\), both give the same tangent line passing through \((1,2)\).
Therefore, the number of tangents is \[ 1 \]
Step 6: Final conclusion.
Hence, the required number of tangents is \[ \boxed{1} \]
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