Chapter 10 — CONIC SECTIONS

Theory Used

The standard form of a parabola opening towards the positive \(x\)-axis is: \[ y^2 = 4ax \]

• Vertex: \((0,0)\)
• Focus: \((a,0)\)
• Directrix: \(x = -a\)
• Axis: \(y = 0\)
• Length of latus rectum: \(4a\)

Solution Roadmap

• Compare given equation with standard form \(y^2 = 4ax\)
• Find parameter \(a\)
• Apply standard results to obtain required elements

Solution

Given: \[ y^2 = 12x \]

Comparing with \(y^2 = 4ax\), we get: \[ 4a = 12 \Rightarrow a = 3 \]

Focus: \((3,0)\)

Directrix: \(x = -3\)

Axis: \(y = 0\)

Length of latus rectum: \(12\)

Geometric Illustration

Exam Significance

• Frequently asked in CBSE Board exams for direct parameter identification
• Core concept for JEE/NEET coordinate geometry problems
• Helps in quick recognition of parabola orientation in MCQs
• Forms the base for advanced topics like tangents, normals, and locus problems

Theory Used

The standard equation of a parabola opening upwards is: \[ x^2 = 4ay \]

• Vertex: \((0,0)\)
• Focus: \((0,a)\)
• Directrix: \(y = -a\)
• Axis: \(x = 0\)
• Length of latus rectum: \(4a\)

Solution Roadmap

• Compare given equation with standard form \(x^2 = 4ay\)
• Determine the value of \(a\)
• Apply standard results to find focus, axis, directrix, and latus rectum

Solution

Given: \[ x^2 = 6y \]

Comparing with \(x^2 = 4ay\), we get: \[ 4a = 6 \Rightarrow a = \dfrac{3}{2} \]

Focus: \(\left(0, \dfrac{3}{2}\right)\)

Axis: \(x = 0\)

Directrix: \(y = -\dfrac{3}{2}\)

Length of latus rectum: \(6\)

Geometric Illustration

Exam Significance

• Standard identification problem frequently asked in CBSE Board exams
• Very common in JEE/NEET for quick parameter extraction
• Helps in recognizing vertical parabolas instantly in MCQs
• Concept is heavily used in tangent, normal, and locus-based problems

Theory Used

The standard equation of a parabola whose axis is along the \(x\)-axis is: \[ y^2 = 4ax \]

• Vertex: \((0,0)\)
• Focus: \((a,0)\)
• Directrix: \(x = -a\)
• Axis: \(y = 0\)
• Length of latus rectum: \(|4a|\)

Solution Roadmap

• Rewrite the given equation in the form \(y^2 = 4ax\)
• Identify the value of \(a\)
• Determine direction of opening (sign of \(a\))
• Apply standard results to find focus, axis, directrix, and latus rectum

Solution

The given equation is: \[ y^2 = -8x \]

The standard form is: \[ y^2 = 4ax \]

Rewriting \(-8x\) as:

\[ -8x = 4 \cdot (-2)x \]

Substituting: \[ y^2 = 4(-2)x \]

Comparing with \(y^2 = 4ax\), we get: \[ a = -2 \]

Since \(a < 0\), the parabola opens towards the negative \(x\)-direction.

Vertex: \[ (0,0) \]

Focus: \[ (a,0) = (-2,0) \]

Axis: \[ y = 0 \]

Directrix: \[ x = -a = 2 \]

Length of latus rectum:

\[ \begin{aligned} |4a| &= |4 \times (-2)| \ &= 8 \end{aligned} \]

Geometric Illustration

Exam Significance

• Very common conceptual question to test understanding of sign of \(a\)
• Frequently appears in JEE/NEET MCQs where direction of opening matters
• Helps avoid common mistake: forgetting that negative \(a\) reverses direction
• Important for solving reflection and tangent-based problems

Theory Used

Standard form: \[ x^2 = 4ay \]

• Vertex: \((0,0)\)
• Focus: \((0,a)\)
• Directrix: \(y = -a\)
• Axis: \(x = 0\)
• Latus rectum length: \(|4a|\)

Solution Roadmap

• Compare with \(x^2 = 4ay\)
• Find \(a\)
• Identify direction using sign of \(a\)
• Apply formulas

Solution

\[ x^2 = -16y \]

\[ x^2 = 4(-4)y \Rightarrow a = -4 \]

Since \(a < 0\), the parabola opens downward.

Vertex: \((0,0)\)

Focus: \[ (0,a) = (0,-4) \]

Axis: \(x = 0\)

Directrix: \[ y = -a = 4 \]

Length of latus rectum:

\[ |4a| = |4 \times (-4)| = 16 \]

Geometric Illustration

Exam Significance

• Tests understanding of sign of \(a\) (direction of opening)
• Very common in CBSE and JEE MCQs
• Helps in quick graph recognition
• Important for solving reflection and tangent problems

Theory Used

Standard form of a parabola opening towards the positive \(x\)-axis: \[ y^2 = 4ax \]

• Vertex: \((0,0)\)
• Focus: \((a,0)\)
• Directrix: \(x = -a\)
• Axis: \(y = 0\)
• Length of latus rectum: \(4a\)

Solution Roadmap

• Compare with \(y^2 = 4ax\)
• Find \(a\)
• Use standard results to determine required elements

Solution

Given: \[ y^2 = 10x \]

Rewriting: \[ 10x = 4 \cdot \dfrac{5}{2} x \]

Comparing with \(y^2 = 4ax\), we get: \[ a = \dfrac{5}{2} \]

Vertex: \((0,0)\)

Focus: \[ \left(\dfrac{5}{2}, 0\right) \]

Axis: \(y = 0\)

Directrix: \[ x = -\dfrac{5}{2} \]

Length of latus rectum:

\[ \begin{aligned} 4a &= 4 \cdot \dfrac{5}{2} \ &= 10 \end{aligned} \]

Geometric Illustration

Exam Significance

• Direct formula application — common in CBSE board exams
• Helps identify right-opening parabolas instantly in MCQs
• Important for solving focus-directrix based problems in JEE/NEET
• Foundation for tangent and normal problems

Theory Used

The standard equation of a parabola whose axis is along the \(y\)-axis is: \[ x^2 = 4ay \]

• Vertex: \((0,0)\)
• Focus: \((0,a)\)
• Directrix: \(y = -a\)
• Axis: \(x = 0\)
• Length of latus rectum: \(|4a|\)

Solution Roadmap

• Compare the given equation with \(x^2 = 4ay\)
• Find the value of \(a\)
• Identify direction using sign of \(a\)
• Apply standard formulas

Solution

Given: \[ x^2 = -9y \]

Rewriting:

\[ -9y = 4 \cdot \left(-\dfrac{9}{4}\right) y \]

\[ x^2 = 4 \left(-\dfrac{9}{4}\right) y \]

Comparing with \(x^2 = 4ay\), we get: \[ a = -\dfrac{9}{4} \]

Since \(a < 0\), the parabola opens downward along the negative \(y\)-axis.

Vertex: \[ (0,0) \]

Focus: \[ \left(0, -\dfrac{9}{4}\right) \]

Axis: \[ x = 0 \]

Directrix: \[ y = -a = \dfrac{9}{4} \]

Length of latus rectum:

\[ \begin{aligned} |4a| &= \left|4 \times \left(-\dfrac{9}{4}\right)\right| \ &= 9 \end{aligned} \]

Geometric Illustration

Exam Significance

• Very common CBSE board question (2–3 marks)
• Tests understanding of sign of \(a\) and direction of opening
• Important for JEE/NEET MCQs involving graph identification
• Foundation for solving tangents, normals, and locus problems

Theory Used

A parabola is the locus of a point which moves such that its distance from a fixed point (focus) is equal to its distance from a fixed line (directrix).

For a parabola with vertex at origin and axis along the \(x\)-axis: \[ y^2 = 4ax \]

• Focus: \((a,0)\)
• Directrix: \(x = -a\)
• Axis: \(y = 0\)

Solution Roadmap

• Identify orientation using focus and directrix
• Locate vertex (midpoint between focus and directrix)
• Find parameter \(a\)
• Write equation using standard form

Solution

Given:

Focus: \((6,0)\), Directrix: \(x = -6\)

The vertex lies midway between the focus and the directrix along the axis.

Since focus is at \(x = 6\) and directrix is \(x = -6\), the midpoint is: \[ \frac{6 + (-6)}{2} = 0 \]

Hence, vertex is: \[ (0,0) \]

The axis is along the \(x\)-axis and the parabola opens towards the positive \(x\)-direction.

Comparing focus \((6,0)\) with \((a,0)\), we get: \[ a = 6 \]

Substituting in standard equation:

\[ \begin{aligned} y^2 &= 4ax \ &= 4 \times 6x \ &= 24x \end{aligned} \]

Hence, the required equation is: \[ y^2 = 24x \]

Geometric Illustration

Exam Significance

• Very important for CBSE Board (3–4 mark descriptive question)
• Tests understanding of focus-directrix definition
• Common in JEE/NEET for identifying equation from geometric data
• Forms base for advanced problems involving shifting of origin

Theory Used

A parabola is the locus of a point equidistant from a fixed point (focus) and a fixed line (directrix).

For a parabola with vertex at origin and axis along the \(y\)-axis: \[ x^2 = 4ay \]

• Focus: \((0,a)\)
• Directrix: \(y = -a\)
• Axis: \(x = 0\)

Solution Roadmap

• Identify orientation from focus and directrix
• Locate vertex (midpoint between focus and directrix)
• Determine value of \(a\)
• Substitute in standard equation

Solution

Given:

Focus: \((0,-3)\), Directrix: \(y = 3\)

The vertex lies midway between the focus and the directrix along the axis.

\[ \frac{-3 + 3}{2} = 0 \]

Hence, vertex: \[ (0,0) \]

Since the focus lies below the vertex and the directrix lies above, the parabola opens downward along the negative \(y\)-axis.

Comparing focus \((0,a)\) with \((0,-3)\), we get: \[ a = -3 \]

Substituting in standard equation:

\[ \begin{aligned} x^2 &= 4ay \ &= 4 \times (-3)y \ &= -12y \end{aligned} \]

Hence, the required equation is: \[ x^2 = -12y \]

Geometric Illustration

Exam Significance

• Frequently asked in CBSE Board exams (3–4 marks)
• Tests understanding of vertical parabolas and sign of \(a\)
• Common in JEE/NEET MCQs involving focus-directrix conversion
• Helps in quick identification of upward vs downward opening curves

Theory Used

A parabola is defined as the locus of a point equidistant from a fixed point (focus) and a fixed line (directrix).

For a parabola with vertex at origin and axis along the \(x\)-axis: \[ y^2 = 4ax \]

• Vertex: \((0,0)\)
• Focus: \((a,0)\)
• Directrix: \(x = -a\)
• Axis: \(y = 0\)

Solution Roadmap

• Identify orientation using focus
• Compare with standard form
• Find value of \(a\)
• Substitute into equation

Solution

Given:

Vertex: \((0,0)\), Focus: \((3,0)\)

Since the focus lies on the positive \(x\)-axis, the axis of the parabola is the \(x\)-axis and the parabola opens towards the positive \(x\)-direction.

Comparing focus \((a,0)\) with \((3,0)\), we get: \[ a = 3 \]

Substituting into standard equation:

\[ \begin{aligned} y^2 &= 4ax \\ &= 4 \times 3x \\ &= 12x \end{aligned} \]

Hence, the required equation is: \[ y^2 = 12x \]

Geometric Illustration

Exam Significance

• Very common CBSE Board question (2–3 marks)
• Tests direct application of standard form
• Important for quick identification in JEE/NEET MCQs
• Foundation for problems involving shifted parabolas

Theory Used

A parabola is the locus of a point equidistant from a fixed point (focus) and a fixed line (directrix).

For a parabola with vertex at origin and axis along the \(x\)-axis: \[ y^2 = 4ax \]

• Vertex: \((0,0)\)
• Focus: \((a,0)\)
• Directrix: \(x = -a\)
• Axis: \(y = 0\)

Solution Roadmap

• Identify orientation using focus
• Compare with standard form
• Determine value of \(a\)
• Substitute into equation

Solution

Given:

Vertex: \((0,0)\), Focus: \((-2,0)\)

Since the focus lies on the negative \(x\)-axis, the parabola opens towards the negative \(x\)-direction. Hence, the axis is the \(x\)-axis.

Comparing focus \((a,0)\) with \((-2,0)\), we get: \[ a = -2 \]

Substituting into standard equation:

\[ \begin{aligned} y^2 &= 4ax \\ &= 4 \times (-2)x \\ &= -8x \end{aligned} \]

Hence, the required equation is: \[ y^2 = -8x \]

Geometric Illustration

Exam Significance

• Common CBSE Board question (2–3 marks)
• Tests understanding of negative \(a\) and direction of opening
• Important for JEE/NEET graph-based MCQs
• Helps avoid common mistake: wrong sign in equation

Theory Used

For a parabola with vertex at origin and axis along the \(x\)-axis, the standard equation is: \[ y^2 = 4ax \]

• Vertex: \((0,0)\)
• Axis: \(y = 0\)
• Focus: \((a,0)\)
• Directrix: \(x = -a\)

Solution Roadmap

• Write standard equation using given axis and vertex
• Substitute the given point to find \(a\)
• Substitute \(a\) back to get final equation

Solution

Given:

Vertex: \((0,0)\), Axis: \(x\)-axis

Hence, the standard form is: \[ y^2 = 4ax \]

Since the parabola passes through \((2,3)\), substituting \(x = 2\), \(y = 3\):

\[ \begin{aligned} y^2 &= 4ax \\ 3^2 &= 4a \cdot 2 \\ 9 &= 8a \\ a &= \dfrac{9}{8} \end{aligned} \]

Substituting \(a = \dfrac{9}{8}\) into the equation:

\[ \begin{aligned} y^2 &= 4ax \\ y^2 &= 4 \cdot \dfrac{9}{8} x \\ y^2 &= \dfrac{9}{2}x \end{aligned} \]

Multiplying both sides by 2:

\[ 2y^2 = 9x \]

Hence, the required equation is: \[ 2y^2 = 9x \]

Geometric Illustration

Exam Significance

• Common CBSE Board question involving parameter determination
• Tests substitution technique (very important for exams)
• Frequently appears in JEE/NEET as MCQ or short answer
• Builds foundation for locus and parameter-based problems

Theory Used

For a parabola with vertex at origin and symmetric about the \(y\)-axis, the standard equation is: \[ x^2 = 4ay \]

• Vertex: \((0,0)\)
• Axis: \(x = 0\)
• Focus: \((0,a)\)
• Directrix: \(y = -a\)

Solution Roadmap

• Use symmetry to identify standard form
• Substitute given point to find \(a\)
• Substitute \(a\) into equation

Solution

Given:

Vertex: \((0,0)\), Symmetry about \(y\)-axis

Hence, the standard equation is: \[ x^2 = 4ay \]

Since the parabola passes through \((5,2)\), substitute \(x=5\), \(y=2\):

\[ \begin{aligned} x^2 &= 4ay \\ 5^2 &= 4a \cdot 2 \\ 25 &= 8a \\ a &= \dfrac{25}{8} \end{aligned} \]

Substituting \(a = \dfrac{25}{8}\):

\[ \begin{aligned} x^2 &= 4ay \ x^2 &= 4 \cdot \dfrac{25}{8} y \\ x^2 &= \dfrac{25}{2} y \end{aligned} \]

Multiplying both sides by 2:

\[ 2x^2 = 25y \]

Hence, the required equation is: \[ 2x^2 = 25y \]

Geometric Illustration

Exam Significance

• Important CBSE Board question (parameter finding)
• Tests understanding of symmetry and standard forms
• Common in JEE/NEET for substitution-based problems
• Builds foundation for coordinate geometry applications